gym 101986

队友写的。好像水水的。

//#pragma GCC optimize(2)

//#pragma GCC optimize(3)

//#pragma GCC optimize(4)

//#pragma GCC optimize("unroll-loops")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC optimize("Ofast,no-stack-protector")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include<bits/stdc++.h>

#define fi first

#define se second

#define db double

#define mp make_pair

#define pb push_back

#define pi acos(-1.0)

#define ll long long

#define vi vector<int>

#define mod 998244353

#define ld long double

//#define C 0.5772156649

//#define ls l,m,rt<<1

//#define rs m+1,r,rt<<1|1

#define pll pair<ll,ll>

#define pil pair<int,ll>

#define pli pair<ll,int>

#define pii pair<int,int>

#define ull unsigned long long

//#define base 1000000000000000000

#define fin freopen("a.txt","r",stdin)

#define fout freopen("a.txt","w",stdout)

#define fio ios::sync_with_stdio(false);cin.tie(0)

inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

inline void sub(ll &a,ll b){a-=b;if(a<)a+=mod;}

inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}

template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}

template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}

inline ll qp(ll a,ll b){ll ans=;while(b){if(b&)ans=ans*a%mod;a=a*a%mod,b>>=;}return ans;}

inline ll qp(ll a,ll b,ll c){ll ans=;while(b){if(b&)ans=ans*a%c;a=a*a%c,b>>=;}return ans;}

using namespace std;

const ull ba=;

const db eps=1e-;

const ll INF=0x3f3f3f3f3f3f3f3f;

const int N=+,maxn=+,inf=0x3f3f3f3f;

ll dp[][];

int main()

{

    int l,k;scanf("%d%d",&l,&k);

    dp[][]=dp[k][]=;

    for(int i=;i<=l;i++)

    {

        dp[i+][]+=dp[i][];

        dp[i+][]+=dp[i][];

        dp[i+k][]+=dp[i][];

    }

    ll ans=;

    for(int i=;i<=l;i++)ans+=dp[i][];

    printf("%lld\n",ans);

    return ;

}

/********************

********************/

C - Medical Checkup

队友写的。

//#pragma GCC optimize(2)

//#pragma GCC optimize(3)

//#pragma GCC optimize(4)

//#pragma GCC optimize("unroll-loops")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC optimize("Ofast,no-stack-protector")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include<bits/stdc++.h>

#define fi first

#define se second

#define db double

#define mp make_pair

#define pb push_back

#define pi acos(-1.0)

#define ll long long

#define vi vector<int>

#define mod 998244353

#define ld long double

//#define C 0.5772156649

//#define ls l,m,rt<<1

//#define rs m+1,r,rt<<1|1

#define pll pair<ll,ll>

#define pil pair<int,ll>

#define pli pair<ll,int>

#define pii pair<int,int>

#define ull unsigned long long

//#define base 1000000000000000000

#define fin freopen("a.txt","r",stdin)

#define fout freopen("a.txt","w",stdout)

#define fio ios::sync_with_stdio(false);cin.tie(0)

inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

inline void sub(ll &a,ll b){a-=b;if(a<)a+=mod;}

inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}

template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}

template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}

inline ll qp(ll a,ll b){ll ans=;while(b){if(b&)ans=ans*a%mod;a=a*a%mod,b>>=;}return ans;}

inline ll qp(ll a,ll b,ll c){ll ans=;while(b){if(b&)ans=ans*a%c;a=a*a%c,b>>=;}return ans;}

using namespace std;

const ull ba=;

const db eps=1e-;

const ll INF=0x3f3f3f3f3f3f3f3f;

const int N=+,maxn=+,inf=0x3f3f3f3f;

ll a[N];

int main()

{

    ll n,t;scanf("%lld%lld",&n,&t);

    for(int i=;i<=n;i++)scanf("%lld",&a[i]);

    ll ma=,sum=;

    for(int i=;i<=n;i++)

    {

        ma=max(ma,a[i]);

        sum=sum+a[i];

        if(t<sum)printf("%d\n",);

        else

        {

            ll te=(t-sum)/ma+;

            printf("%lld\n",te);

        }

    }

    return ;

}

/********************

********************/

B - Parallel Lines

dfs暴力枚举两两组合的情况，枚举选第一个没被选的和枚举的组合，这样能把复杂度降到最低。

然后答案一遍加入一遍更新，好像我常数扣的有点厉害。复杂度是15 * 13 * 11 * ... * 3

#include<bits/stdc++.h>

#define LL long long

#define fi first

#define se second

#define mk make_pair

#define PLL pair<LL, LL>

#define PLI pair<LL, int>

#define PII pair<int, int>

#define SZ(x) ((int)x.size())

#define ull unsigned long long

using namespace std;

const int N =  + ;

const int M =  + ;

const int inf = 0x3f3f3f3f;

const LL INF = 0x3f3f3f3f3f3f3f3f;

const int mod = 1e9 + ;

const double eps = 1e-;

const double PI = acos(-);

struct Point {

    int x, y;

} p[N];

int X[][], Y[][], all;

int num[][];

int n, cnt[ << ], who[ << ];

int ans, tmp;

vector<int> go[ << ];

inline void change(int x, int y, int op) {

    x += , y += ;

    if(op == ) tmp += num[x][y];

    else tmp -= num[x][y] - ;

    num[x][y] += op;

}

void dfs(int S) {

    if(cnt[S] +  > n) {

        ans = max(ans, tmp);

        return;

    }

    for(int i = ; i < SZ(go[S]); i++) {

        change(X[go[S][]][go[S][i]], Y[go[S][]][go[S][i]], );

        dfs(S | ( << go[S][]) | ( << go[S][i]));

        change(X[go[S][]][go[S][i]], Y[go[S][]][go[S][i]], -);

    }

}

int main() {

    for(int i = ; i < ( << ); i++)

        cnt[i] = cnt[i - (i & -i)] + ;

    scanf("%d", &n);

    all = n / ;

    for(int i = ; i < n; i++) {

        scanf("%d%d", &p[i].x, &p[i].y);

    }

    for(int i = ; i < n; i++) {

        for(int j = i + ; j < n; j++) {

            X[i][j] = p[i].x - p[j].x;

            Y[i][j] = p[i].y - p[j].y;

            int gcd = __gcd(X[i][j], Y[i][j]);

            X[i][j] /= gcd;

            Y[i][j] /= gcd;

            if(X[i][j] < ) X[i][j] = -X[i][j], Y[i][j] = -Y[i][j];

//            printf("%d %d: %d %d\n", i, j, X[i][j], Y[i][j]);

        }

    }

    for(int i = ; i < ( << n); i++)

        for(int k = ; k < n; k++)

            if(!(i >> k & )) go[i].push_back(k);

    dfs();

    printf("%d\n", ans);

    return ;

}

/*

*/

I - Starting a Scenic Railroad Service

队友写的。

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<vector>

#include<queue>

#include<iostream>

#include<stack>

#include<string>

#include<map>

using namespace std;

#define pb push_back

#define pf push_front

#define lb lower_bound

#define ub upper_bound

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define mst(x,a) memset(x,a,sizeof(x))

#define all(x) (x).begin(),(x).end()

#define CLOSE ios::sync_with_stdio(false)

typedef pair<int,int> pii;

typedef long long ll;

typedef vector<int> vi;

#define fi first

#define se second

#define sz(x) ((int)x.size())

#define cl(x) x.clear()

const int mod = ;

const int N = ;

const int INF=0x3f3f3f3f;

void MOD(ll &a){if(a>=mod) a-=mod;}

void MOD(ll &a,ll c){if(a>=c) a-=c;}

void ADD(ll &a,ll b){ a+=b; MOD(a);}

void ADD(ll &a,ll b,ll c){a+=b;MOD(a,c);}

ll qpow(ll a,ll b){ll ans=;while(b){if(b&)ans=ans*a%mod;a=a*a%mod;b/=;}return ans;}

ll qpow(ll a,ll b,ll c){ll ans=;while(b){if(b&)ans=ans*a%c;a=a*a%c;b/=;}return ans;}

int n;

int x[N],y[N],B[N];

struct seg{

    int l,r;

}a[N],b[N];

bool cmp1(seg a,seg b){

    if(a.l==b.l) return a.r<b.r;

    return a.l<b.l;

}

bool cmp2(seg a,seg b){

    if(a.r==b.r) return a.l<b.l;

    return a.r<b.r;

}

int main(){

    scanf("%d",&n);

    for(int i=;i<=n;i++){

        scanf("%d%d",&x[i],&y[i]);

        y[i]--;

        B[x[i]]++; B[y[i]+]--;

        a[i].l=x[i]; a[i].r=y[i];

        b[i].l=x[i]; b[i].r=y[i];

    }

    sort(a+,a++n,cmp1);

    sort(b+,b++n,cmp2);

    int ans1=,ans2=;

    for(int i=;i<=;i++){

        B[i]+=B[i-]; ans2=max(ans2,B[i]);

    }

    //a sort by l    b sort by r

    for(int i=;i<=n;i++){

        int tmp=n;

        int l=,r=n+;

        while(l+<r){

            int m=l+r>>;

            if(a[m].l>y[i]) r=m;

            else l=m;

        }

        tmp-=n-r+;

        l=,r=n;

        while(l+<r){

            int m=l+r>>;

            if(b[m].r<x[i]) l=m;

            else r=m;

        }

        tmp-=l;

        ans1=max(ans1,tmp);

    }

    printf("%d %d\n",ans1,ans2);

}

/*

*/

F - Pizza Delivery

感觉是比较常规的套路题。

把边(u, v, w)反转以后如果 dis[ 1 ][ v ] + dis[ 2 ][ u ] + w < 最短路则是happy，这里有个问题就是反转之后1到不了v，但是反转前到到的了，

那么dis[ 1 ][ v ] 就不是INF，但是通过分析我们能发现这不影响结果。

然后就是 sad 和 soso，这个对 1 到 2 之间的最短路图求个桥就好了，看看反转的是不是桥。

#include<bits/stdc++.h>

#define LL long long

#define fi first

#define se second

#define mk make_pair

#define PLL pair<LL, LL>

#define PLI pair<LL, int>

#define PII pair<int, int>

#define SZ(x) ((int)x.size())

#define ull unsigned long long

using namespace std;

const int N = 1e5 + ;

const int inf = 0x3f3f3f3f;

const LL INF = 0x3f3f3f3f3f3f3f3f;

const int mod = 1e9 + ;

const double eps = 1e-;

const double PI = acos(-);

int n, m, a[N], b[N], c[N];

LL dis[][N];

bool flag[N];

struct node {

    LL d;

    int to, id;

    bool operator < (const node& rhs) const {

        return d > rhs.d;

    }

};

vector<node> G[N], rG[N], sG[N];

priority_queue<node> que;

void Dij(int S, LL dis[N]) {

    dis[S] = ;

    que.push(node{, S, });

    while(!que.empty()) {

        int u = que.top().to;

        LL d = que.top().d; que.pop();

        if(d > dis[u]) continue;

        for(auto& t : G[u]) {

            if(d + t.d < dis[t.to]) {

                dis[t.to] = d + t.d;

                que.push(node{dis[t.to], t.to, });

            }

        }

    }

}

void Dij2(int S, LL dis[N]) {

    dis[S] = ;

    que.push(node{, S, });

    while(!que.empty()) {

        int u = que.top().to;

        LL d = que.top().d; que.pop();

        if(d > dis[u]) continue;

        for(auto& t : rG[u]) {

            if(d + t.d < dis[t.to]) {

                dis[t.to] = d + t.d;

                que.push(node{dis[t.to], t.to, });

            }

        }

    }

}

int dfn[N], low[N], idx;

bool in[N];

void tarjan(int u, int fa) {

    in[u] = true; ++idx;

    dfn[u] = low[u] = idx;

    for(node& t : sG[u]) {

        int v = t.to;

        if(v == fa) continue;

        if(!dfn[v]) {

            tarjan(v, u);

            low[u] = min(low[u], low[v]);

            if(dfn[u] < low[v]) flag[t.id] = true;

        } else if(in[v]) {

            low[u] = min(low[u], dfn[v]);

        }

    }

    in[u] = false;

}

int main() {

    memset(dis, INF, sizeof(dis));

    scanf("%d%d", &n, &m);

    for(int i = ; i <= m; i++) {

        scanf("%d%d%d", &a[i], &b[i], &c[i]);

        G[a[i]].push_back(node{c[i], b[i], i});

        rG[b[i]].push_back(node{c[i], a[i], i});

    }

    Dij(, dis[]);

    Dij2(, dis[]);

    LL mn = dis[][];

    if(mn != INF) {

        for(int i = ; i <= n; i++) {

            for(node t : G[i]) {

                if(t.d + dis[][i] + dis[][t.to] == mn) {

                    sG[i].push_back(t);

                    sG[t.to].push_back(node{t.d, i, t.id});

                }

            }

        }

        tarjan(, );

        for(int i = ; i <= m; i++) {

            if(dis[][b[i]] + dis[][a[i]] + c[i] < mn) {

                puts("HAPPY");

            } else {

                if(flag[i]) puts("SAD");

                else puts("SOSO");

            }

        }

    } else {

        for(int i = ; i <= m; i++) {

            if(dis[][b[i]] + dis[][a[i]] + c[i] < mn) {

                puts("HAPPY");

            } else {

                puts("SAD");

            }

        }

    }

    return ;

}

/*

*/

赛后补题**********************************************************************************************************************************************

G - Rendezvous on a Tetrahedron

训练结束半个多小时才码出来。。。把四面体展开，然后暴力的取找到它最后的位置在哪里，好像有更方便的方法，我写得好麻烦啊啊。。

#include<bits/stdc++.h>

#define LL long long

#define fi first

#define se second

#define mk make_pair

#define PLL pair<LL, LL>

#define PLI pair<LL, int>

#define PII pair<int, int>

#define SZ(x) ((int)x.size())

#define ull unsigned long long

using namespace std;

const int N = 1e5 + ;

const int inf = 0x3f3f3f3f;

const LL INF = 0x3f3f3f3f3f3f3f3f;

const int mod = 1e9 + ;

const double eps = 1e-;

const double PI = acos(-);

struct Point {

    double x, y;

    void print() {

        printf("%.3f %.3f ^^\n", x, y);

    }

    Point(double x = , double y = ) : x(x), y(y) { }

};

typedef Point Vector;

int dcmp(double x) {

    if(fabs(x) < eps) return ;

    else return x <  ? - : ;

}

Point operator + (Vector A, Vector B) {return Point(A.x + B.x, A.y + B.y);}

Point operator - (Vector A, Vector B) {return Point(A.x - B.x, A.y - B.y);}

Point operator * (Vector A, double p) {return Point(A.x * p, A.y * p);}

Point operator / (Vector A, double p) {return Point(A.x / p, A.y / p);}

bool operator < (const Vector &A, const Vector &B) {return A.y < B.y || (A.y == B.y && A.x < B.x);}

bool operator == (const Vector &A, const Point &B) {return dcmp(A.x - B.x) ==  && dcmp(A.y - B.y) == ;}

double Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;}

double Length(Vector A) {return sqrt(Dot(A, A));}

double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));}

double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;}

double Area2(Point A, Point B, Point C) {return Cross(B - A, C - A);}

Vector Rotate(Vector A, double rad) {

    return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));

}

double GetLineIntersectionTime(Point P, Vector v, Point Q, Vector w) {

    Vector u = P - Q;

    double t = Cross(w, u) / Cross(v, w);

    return t;

}

double dist(const Point& a, const Point &b) {

    return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));

}

bool isPointOnSegment(const Point &p, const Point &a1, const Point &a2)

{

    if(dcmp(Cross(a1-p,a2-p))) return ;

    else if(dcmp(p.x-min(a1.x,a2.x))>=&&dcmp(p.x-max(a1.x,a2.x))<=

            &&dcmp(p.y-min(a1.y,a2.y))>=&&dcmp(p.y-max(a1.y,a2.y))<=) return ;

    else return ;

}

int isPointInPolygon(Point p, Point *poly, int n) {

    int wn = ;

    for(int i = ; i < n; i++) {

        if(isPointOnSegment(p, poly[i], poly[(i+)%n])) return -; //在边界上

        int k = dcmp(Cross(poly[(i+)%n]-poly[i], p-poly[i]));

        int d1 = dcmp(poly[i].y-p.y);

        int d2 = dcmp(poly[(i+)%n].y-p.y);

        if(k> && d1<= && d2>) wn++;

        if(k< && d2<= && d1>) wn--;

    }

    if(wn != ) return ; //内部

    return ; //外部

}

Point A2 = Point(, );

Point A3 = Point(, );

Point A1 = Point(, sqrt(3.0));

Point D = (A1 + A2) / ;

Point B = (A1 + A3) / ;

Point C = (A2 + A3) / ;

Point poly1[] = {A1, D, B};

Point poly2[] = {D, C, B};

Point poly3[] = {D, A2, C};

Point poly4[] = {B, C, A3};

bool isA12(Point x) {

    return isPointOnSegment(x, A1, A2);

}

bool isA23(Point x) {

    return isPointOnSegment(x, A2, A3);

}

bool isA31(Point x) {

    return isPointOnSegment(x, A3, A1);

}

int belong(Point x) {

    if(isPointInPolygon(x, poly1, ) == ) return ;

    if(isPointInPolygon(x, poly2, ) == ) return ;

    if(isPointInPolygon(x, poly3, ) == ) return ;

    if(isPointInPolygon(x, poly4, ) == ) return ;

}

Point dfs(Point p, Vector v, double len) {

    if(isA12(p)) {

        double t1 = GetLineIntersectionTime(p, v, A2, A3 - A2);

        double t2 = GetLineIntersectionTime(p, v, A1, A3 - A1);

        Point nxtp;

        if(dcmp(t1) >  && (dcmp(t2) <=  || dcmp(t1) >  && t1 < t2)) {

            nxtp = p + (v * t1);

            double gg = dist(p, nxtp);

            if(dcmp(len - gg) < ) {

                return p + (v / Length(v)) * len;

            } else {

                nxtp = C * 2.0 - nxtp;

                v = Rotate(v, PI);

                return dfs(nxtp, v, len - gg);

            }

        } else {

            nxtp = p + v * t2;

            double gg = dist(p, nxtp);

            if(dcmp(len - gg) < ) {

                return p + (v / Length(v)) * len;

            } else {

                nxtp = B * 2.0 - nxtp;

                v = Rotate(v, PI);

                return dfs(nxtp, v, len - gg);

            }

        }

    } else if(isA23(p)) {

        double t1 = GetLineIntersectionTime(p, v, A1, A1 - A2);

        double t2 = GetLineIntersectionTime(p, v, A1, A1 - A3);

        Point nxtp;

        if(dcmp(t1) >  && (dcmp(t2) <=  || dcmp(t1) >  && t1 < t2)) {

            nxtp = p + v * t1;

            double gg = dist(p, nxtp);

            if(dcmp(len - gg) < ) {

                return p + (v / Length(v)) * len;

            } else {

                nxtp = D * 2.0 - nxtp;

                v = Rotate(v, PI);

                return dfs(nxtp, v, len - gg);

            }

        } else {

            nxtp = p + v * t2;

            double gg = dist(p, nxtp);

            if(dcmp(len - gg) < ) {

                return p + (v / Length(v)) * len;

            } else {

                nxtp = B * 2.0 - nxtp;

                v = Rotate(v, PI);

                return dfs(nxtp, v, len - gg);

            }

        }

    } else if(isA31(p)) {

        double t1 = GetLineIntersectionTime(p, v, A2, A2 - A1);

        double t2 = GetLineIntersectionTime(p, v, A2, A2 - A3);

        Point nxtp;

        if(dcmp(t1) >  && (dcmp(t2) <=  || dcmp(t1) >  && t1 < t2)) {

            nxtp = p + v * t1;

            double gg = dist(p, nxtp);

            if(dcmp(len - gg) < ) {

                return p + (v / Length(v)) * len;

            } else {

                nxtp = D * 2.0 - nxtp;

                v = Rotate(v, PI);

                return dfs(nxtp, v, len - gg);

            }

        } else {

            nxtp = p + v * t2;

            double gg = dist(p, nxtp);

            if(dcmp(len - gg) < ) {

                return p + (v / Length(v)) * len;

            } else {

                nxtp = C * 2.0 - nxtp;

                v = Rotate(v, PI);

                return dfs(nxtp, v, len - gg);

            }

        }

    }

}

Point read() {

    char s[];

    double d, l;

    scanf("%s%lf%lf", s, &d, &l);

    if(s[] == 'C' && s[] == 'D') {

        Point startp = A2;

        Vector startv = Rotate(C - A2, d /  * PI);

        return dfs(startp, startv, l);

    } else if(s[] == 'D' && s[] == 'B') {

        Point startp = A1;

        Vector startv = Rotate(D - A1, d /  * PI);

        return dfs(startp, startv, l);

    } else {

        Point startp = A3;

        Vector startv = Rotate(B - A3, d /  * PI);

        return dfs(startp, startv, l);

    }

}

int main() {

    Point Point1 = read();

    Point Point2 = read();

    if(belong(Point1) == belong(Point2)) puts("YES");

    else puts("NO");

    return ;

}

/*

*/

E - Black or White

不会写，但是看了题解觉得好有道理啊啊啊，我怎么没想到呢。

我们把t[ i ] == s[ i ]的位置的颜色保留，把颜色不同的当成没有颜色就好了。

我们定义dp[ i ] 表示把 1 - i 全部染成对应颜色所需要的次数。

我们把 t 中连续相同的一段看成一个整体。变成 WBWBWB的形式，如果长度为 m 则答案为 (m / 2) + 1

如果t[ i ] == s[ i ] dp[ i ] = dp[ i - 1]

否则 dp[ i ] = min(dp[ j ] + (sum[ i ] - sum[ j + 1 ]) / 2 + 1)

很显然能发现能用单调队列优化。

#include<bits/stdc++.h>

#define LL long long

#define fi first

#define se second

#define mk make_pair

#define PLL pair<LL, LL>

#define PLI pair<LL, int>

#define PII pair<int, int>

#define SZ(x) ((int)x.size())

#define ull unsigned long long

using namespace std;

const int N = 5e5 + ;

const int inf = 0x3f3f3f3f;

const LL INF = 0x3f3f3f3f3f3f3f3f;

const int mod = 1e9 + ;

const double eps = 1e-;

const double PI = acos(-);

int n, k, head, rear, sum[N], que[N];

int dp[N];

char s[N], t[N];

int main() {

    scanf("%d%d", &n, &k);

    scanf("%s%s", s + , t + );

    for(int i = ; i <= n; i++) sum[i] = sum[i - ] + (t[i] != t[i - ]);

    head = , rear = ;

    dp[] = ;

    que[++rear] = ;

    for(int i = ; i <= n; i++) {

        if(t[i] == s[i]) {

            dp[i] = dp[i - ];

        } else {

            while(head <= rear && i - que[head] > k) head++;

            dp[i] = dp[que[head]] + (sum[i] - sum[que[head] + ] + ) /  + ;

        }

        while(head <= rear && *dp[i] - sum[i + ] <= *dp[que[rear]] - sum[que[rear] + ]) rear--;

        que[++rear] = i;

    }

    printf("%d\n", dp[n]);

    return ;

}

/*

*/

gym 101986的更多相关文章

ACM: Gym 101047M Removing coins in Kem Kadrãn - 暴力
Gym 101047M Removing coins in Kem Kadrãn Time Limit:2000MS Memory Limit:65536KB 64bit IO Fo ...
ACM: Gym 101047K Training with Phuket's larvae - 思维题
Gym 101047K Training with Phuket's larvae Time Limit:2000MS Memory Limit:65536KB 64bit IO F ...
ACM: Gym 101047E Escape from Ayutthaya - BFS
Gym 101047E Escape from Ayutthaya Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I6 ...
ACM: Gym 101047B Renzo and the palindromic decoration - 手速题
Gym 101047B Renzo and the palindromic decoration Time Limit:2000MS Memory Limit:65536KB 64 ...
Gym 101102J---Divisible Numbers(反推技巧题)
题目链接 http://codeforces.com/gym/101102/problem/J Description standard input/output You are given an a ...
Gym 100917J---Judgement(01背包+bitset)
题目链接 http://codeforces.com/gym/100917/problem/J Description standard input/outputStatements The jury ...
Gym 100917J---dir -C（RMQ--ST）
题目链接 http://codeforces.com/gym/100917/problem/D problem description Famous Berland coder and IT mana ...
Gym 101102D---Rectangles(单调栈)
题目链接 http://codeforces.com/gym/101102/problem/D problem description Given an R×C grid with each cel ...
Gym 101102C---Bored Judge（区间最大值）
题目链接 http://codeforces.com/gym/101102/problem/C problem description Judge Bahosain was bored at ACM ...

随机推荐

MT【75】考察高斯函数的一道高考压轴题
解答:答案1,3,4. 这里关于高斯函数$[x]$的一个不等式是需要知道的$x-1<[x]\le x$,具体的:
【uoj428】普通的计数题
Portal --> uoj428 Solution 不会胖子的一个log正解qwq只能怂怂滴写分治了qwq 首先就是一个我想不到的转化qwq 我们将第$i$次操作加入的数看成一个编 ...
项目实战SportsStore——订单处理模块
前面的步骤如果顺利完成,你的网站运行之后应该能够正常显示下面三个页面: 1.产品列表 2.购物车内容页面在某个商品后面点击“添加到购物车”则出现下图页面: 此页面上点击“继续购物”按钮则返回到产品列 ...
jQueryCDN
分享几个jquery的几个国内国外的CDN加速节点,方便广大的开发设计者调用和节约空间,官网的总是最新版本的jquery所以不用去担心版本更新问题,其他加速节点可能不会在更新版本,所以取舍问题自己决定 ...
浅说搜索引擎和SEO
搜索引擎基本工作原理信息收集功能技术点SEO优化核心优化重点 SEO优化网站URL title信息 meta信息图片ALT flash信息 frame框架网页重要度特征之前有看到一个师 ...
转：IOS 基于APNS消息推送原理与实现(JAVA后台)
Push的原理: Push 的工作机制可以简单的概括为下图图中,Provider是指某个iPhone软件的Push服务器,这篇文章我将使用.net作为Provider. APNS 是Apple ...
IO 复习字节流字符流拷贝文件
/* 本地文件 URL 文件拷贝 *//*文本文件拷贝可以通过字符流,也可以通过字节流*/ /*二进制文件拷贝只可以通过字节流*//* 希望这个例子能帮助搞懂字符流与字节流的区别 */ imp ...
CSS的力量：用一个DIV画图
这些图片都是用一个DIV绘制出来的,其实原理并不复杂. 这些图片都是由CSS绘制出来的,通过background-image叠加实现, 如蘑菇头的实现,通过 radial-gradient 径向渐变 ...
iOS-Socket编程体验
CHENYILONG Blog Socket编程体验 Socket编程体验技术博客http://www.cnblogs.com/ChenYilong/新浪微博http://weibo.com/lu ...
各种initcall的执行先后顺序（module_init、postcore_initcall、arch_initcall、subsys_initcall、 fs_initcall）【转】
转自:http://www.cnblogs.com/superlcc/archive/2012/09/12/2681930.html 现在以module_init为例分析initcall在内核中的调用 ...

gym 101986

gym 101986的更多相关文章

随机推荐

热门专题