$$2015-2016\ ACM-ICPC,\ NEERC,\ Northern\ Subregional\ Contest$$

\(A.Alex\ Origami\ Squares\)

签到

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<bits/stdc++.h>

using namespace std;

int main(){

    int a,b;

    freopen("alex.in","r",stdin);

    cin >> a >> b;

    freopen("alex.out","w",stdout);

    double res = max(max(min(a/3.0,b/1.0),min(a/1.0,b/3.0)),min(a/2.0,b/2.0));

    cout << res << endl;

    return 0;

}

\(B.Black\ and\ White\)

构造，把多的框在少的里面即可

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<bits/stdc++.h>

using namespace std;

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

const int MAXN = 1e5+7;

int n,m,len;

char tile[MAXN][5];

int main(){

    #ifdef ONLINE_JUDGE

    freopen("black.in","r",stdin);

    #endif

    ____();

    cin >> n >> m;

    #ifdef ONLINE_JUDGE

    freopen("black.out","w",stdout);

    #endif

    if(m==0){

        cout << 1 << ' ' << 1 << endl << '@' << endl;

        return 0;

    }

    if(n==0){

        cout << 1 << ' ' << 1 << endl << '.' << endl;

        return 0;

    }

    char x='@',y='.';

    if(n<m){

        swap(n,m);

        swap(x,y);

    }

    len = 1;

    if(n>m){

        tile[len][1] = tile[len][2] = tile[len][3] = y;

        len++, m--;

    }

    while(n>m){

        tile[len][2] = x;

        tile[len][1] = tile[len][3] = y;

        len++, n--;

        tile[len][1] = tile[len][2] = tile[len][3] = y;

        len++;

    }

    while(n){

        tile[len][1] = tile[len][2] = tile[len][3] = x;

        len++, n--;

        tile[len][1] = tile[len][2] = tile[len][3] = y;

        len++;

    }

    if(tile[len][1]=='\0') len--;

    cout << len << ' ' << 3 << endl;

    for(int i = 1; i <= len; i++) cout << (tile[i]+1) << endl;

    return 0;

}

\(C.Concatenation\)

答案就是总数量-右串从\(1\)到\(lenr-1\)各个字符在左串\(2~lenl\)之间出现的次数

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<bits/stdc++.h>

using namespace std;

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

const int MAXN = 3e5+7;

char s[MAXN],t[MAXN];

int len1,len2;

int main(){

    #ifdef ONLINE_JUDGE

    freopen("concatenation.in","r",stdin);

    #endif

    scanf("%s %s",s+1,t+1);

    #ifdef ONLINE_JUDGE

    freopen("concatenation.out","w",stdout);

    #endif

    len1 = strlen(s+1);

    len2 = strlen(t+1);

    int_fast64_t res = len1*(int_fast64_t)len2;

    map<char,int> mp;

    for(int i = 2; i <= len1; i++) mp[s[i]]++;

    for(int i = 1; i < len2; i++) res -= mp[t[i]];

    cout << res << endl;

    return 0;

}

\(D.Distribution\ in\ Metagonia\)

考虑构造时先把质因子\(2\)，\(3\)提出来，然后得到一个奇数，把奇数拆分为\(3^k·p\)，继续对\(p\)拆分直到\(p\)的因子只含\(2\)和\(3\)为止

递归构造即可

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<bits/stdc++.h>

using namespace std;

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

using LL = int_fast64_t;

int t;

LL n[1111];

void gao(LL m, LL mul, vector<LL> &vec){

    if(m==1){

        vec.emplace_back(mul);

        return;

    }

    if(m%3!=0&&m%2!=0){

        LL tp = 1;

        while(tp*3<=m) tp *= 3;

        gao(tp,mul,vec);

        gao(m-tp,mul,vec);

        return;

    }

    LL sum = 1;

    while(m%2==0){

        sum *= 2;

        m >>= 1;

    }

    while(m%3==0){

        sum *= 3;

        m /= 3;

    }

    gao(m,sum*mul,vec);

}

void solve(LL m){

    vector<LL> vec;

    gao(m,1,vec);

    cout << vec.size() << endl;

    for(LL x : vec) cout << x << ' '; cout << endl;

}

int main(){

    #ifdef ONLINE_JUDGE

    freopen("distribution.in","r",stdin);

    freopen("distribution.out","w",stdout);

    #endif

    ____();

    cin >> t;

    for(int i = 1; i <= t; i++) cin >> n[i];

    for(int i = 1; i <= t; i++) solve(n[i]);

    return 0;

}

\(E.Easy\ Arithmetic\)

打记号标记前导零、上一位是否是符号、当前正负值然后判断即可

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<bits/stdc++.h>

using namespace std;

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

const int MAXN = 3333;

char s[MAXN],tar[MAXN];

int main(){

    #ifdef ONLINE_JUDGE

    freopen("easy.in","r",stdin);

    #endif

    ____();

    cin >> (s+1);

    int len = strlen(s+1);

    #ifdef ONLINE_JUDGE

    freopen("easy.out","w",stdout);

    #endif

    int c = 1, ptr = 1;

    bool tag = true, lead = true, op = true;

    while(ptr<=len){

        if(!isdigit(s[ptr])){

            if(s[ptr]=='-') tag = false;

            else if(s[ptr]=='+') tag = true;

            tar[c++] = s[ptr];

            op = true;

        }

        else{

            if(op){

                op = false;

                tar[c++] = s[ptr];

                lead = (s[ptr]=='0');

            }

            else{

                if(!tag){

                    tar[c++] = '+';

                    tag = true;

                    tar[c++] = s[ptr];

                    lead = (s[ptr]=='0');

                }

                else{

                    if(lead){

                        tar[c++] = '+';

                        lead = (s[ptr]=='0');

                        tar[c++] = s[ptr];

                    }

                    else tar[c++] = s[ptr];

                }

            }

        }

        ptr++;

    }

    cout << (tar+1) << endl;

    return 0;

}

\(F.Fygon\)

\(G.Graph\)

友情链接

大顶堆小顶堆维护，做法有点妙

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<bits/stdc++.h>

using namespace std;

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

const int MAXN = 1e5+7;

int n,m,k,pre,ord[MAXN],cur,deg[MAXN];

vector<int> G[MAXN];

set<int> S;

set<int,greater<int>> done;

vector<pair<int,int>> edge;

void toposort(int u){

    ord[++cur] = u;

    pre = u;

    for(int v : G[u]) if(--deg[v]==0) S.insert(v);

}

int main(){

    #ifdef ONLINE_JUDGE

    freopen("graph.in","r",stdin);

    freopen("graph.out","w",stdout);

    #endif

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

    for(int i = 1; i <= m; i++){

        int u, v;

        scanf("%d %d",&u,&v);

        G[u].emplace_back(v);

        deg[v]++;

    }

    for(int i = 1; i <= n; i++) if(deg[i]==0) S.insert(i);

    while(cur<n){

        if(S.empty()){          //自己安排下一个节点顺序

            int u = *done.begin();

            done.erase(u);

            edge.emplace_back(make_pair(pre,u));

            toposort(u);

        }

        else if(!k||(S.size()==1&&(done.empty()||*S.begin()>*done.begin()))){

            int u = *S.begin();

            S.erase(u);

            toposort(u);

        }

        else{

            k--;

            done.insert(*S.begin());

            S.erase(S.begin());

        }

    }

    for(int i = 1; i <= n; i++) printf("%d ",ord[i]); puts("");

    printf("%d\n",edge.size());

    for(auto p : edge) printf("%d %d\n",p.first,p.second);

    return 0;

}

\(H.Hash\ Code\ Hacker\)

每次从后一位向前进一位，后一位\(y\)变成\(Z\)，前一位的\(x\)变成\(y\)，重复操作即可

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<bits/stdc++.h>

using namespace std;

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

const int MAXN = 1111;

char s[MAXN];

int k;

int main(){

    #ifdef ONLINE_JUDGE

    freopen("hash.in","r",stdin);

    #endif

    cin >> k;

    #ifdef ONLINE_JUDGE

    freopen("hash.out","w",stdout);

    #endif

    for(int i = 1; i <= 999; i++) s[i] = 'x';

    s[1000] = 'y';

    cout << (s+1) << endl;

    for(int i = 1; i < k; i++){

        s[1000+1-i] = 'Z';

        s[1000-i] = 'y';

        cout << (s+1) << endl;

    }

    return 0;

}

\(I.Insider's\ Information\)

\(J.Journey\ to\ the\ "The World's Start"\)

二分答案之后线段树维护DP求最小耗时

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<bits/stdc++.h>

using namespace std;

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

const int MAXN = 1e5+7;

using LL = int_fast64_t;

const LL INF = 0x3f3f3f3f3f3f3f3f;

int n,t,p[MAXN],cost[MAXN];

struct SegmentTree{

    LL minn[MAXN<<2];

    int l[MAXN<<2],r[MAXN<<2];

    #define ls(rt) ((rt) << 1)

    #define rs(rt) ((rt) << 1 | 1)

    void pushup(int rt){ minn[rt] = min(minn[ls(rt)],minn[rs(rt)]); }

    void build(int L, int R, int rt){

        l[rt] = L, r[rt] = R;

        if(L+1==R) return;

        int mid = (L+R) >> 1;

        build(L,mid,ls(rt)); build(mid,R,rs(rt));

        pushup(rt);

    }

    void update(int pos, int rt, LL x){

        if(l[rt]+1==r[rt]) {

            minn[rt] = x;

            return;

        }

        int mid = (l[rt]+r[rt]) >> 1;

        if(pos<mid) update(pos,ls(rt),x);

        else update(pos,rs(rt),x);

        pushup(rt);

    }

    LL qmin(int L, int R, int rt){

        if(l[rt]>=R||L>=r[rt]) return INF;

        if(L<=l[rt] && r[rt]<=R) return minn[rt];

        return min(qmin(L,R,ls(rt)),qmin(L,R,rs(rt)));

    }

}ST;

bool check(int mid){

    ST.update(1,1,-1);

    for(int i = 2; i <= n; i++) ST.update(i,1,ST.qmin(max(1,i-mid),i,1)+cost[i]);

    return ST.qmin(n,n+1,1)+n<=t;

}

int solve(){

    int l = 1, r = n-1;

    ST.build(1,n+1,1);

    while(l<=r){

        int mid = (l+r)>>1;

        if(check(mid)) r = mid - 1;

        else l = mid + 1;

    }

    return p[l];

}

int main(){

    #ifdef ONLINE_JUDGE

    freopen("journey.in","r",stdin);

    freopen("journey.out","w",stdout);

    #endif

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

    for(int i = 1; i < n; i++) scanf("%d",&p[i]);

    for(int i = n-2; i >= 1; i--) p[i] = min(p[i],p[i+1]);

    for(int i = 2; i < n; i++) scanf("%d",&cost[i]);

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

    return 0;

}

\(K.Kingdom\ Trip\)

\(L.Lucky\ Chances\)

签到，暴力

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<bits/stdc++.h>

using namespace std;

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

const int MAXN = 111;

const int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};

int n,m,A[MAXN][MAXN];

int main(){

    #ifdef ONLINE_JUDGE

    freopen("lucky.in","r",stdin);

    #endif

    ____();

    cin >> n >> m;

    for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) cin >> A[i][j];

    #ifdef ONLINE_JUDGE

    freopen("lucky.out","w",stdout);

    #endif

    int res = 0;

    for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++){

        for(int d = 0; d < 4; d++){

            bool ok = true;

            for(int x = i+dir[d][0], y = j + dir[d][1]; x>0&&x<=n&&y>0&&y<=m; x+=dir[d][0],y+=dir[d][1]) if(A[x][y]>=A[i][j]) ok = false;

            res+=ok;

        }

    }

    cout << res << endl;

    return 0;

}

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