hdu多校4

Problem L. Graph Theory Homework

思路：很容易想到一步从 1 走到 n 最优。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int,int>
#define piii pair<int, pair<int,int>>

using namespace std;

const int N = 2e5 + ;
const int M = 1e4 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;
const double PI = acos(-);

int n, a[N];
int main() {
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        for(int i = ; i <= n; i++) scanf("%d", &a[i]);
        printf("%d\n", (int)sqrt(abs(a[n] - a[])));
    }
    return ;
}
/*
*/

Problem K. Expression in Memories

思路：把0的情况搞清楚就好啦。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int,int>
#define piii pair<int, pair<int,int>>

using namespace std;

const int N = 2e5 + ;
const int M = 1e4 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;
const double PI = acos(-);

char s[N];
int n;
bool is(char c) {
    if(c == '+' || c == '*') return true;
    return false;
}

int main() {
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%s", s + );
        n = strlen(s + );
        s[] = '+';
        bool flag = true;

        for(int i = ; i <= n; i++) {
            if(is(s[i]) && is(s[i - ])) {
                flag = false;
                break;
            }
        }
        if(is(s[]) || is(s[n])) flag = false;

        if(!flag) {
            puts("IMPOSSIBLE");
            continue;
        }

        for(int i = ; i <= n && flag; i++) {
            if(s[i] == '') {
                if(s[i - ] >= '' && s[i - ] <= '') continue;
                else if(is(s[i - ])) {
                    if(i < n) {
                        if(!is(s[i + ])) {
                            if(s[i + ] == '?') s[i + ] = '+';
                            else {
                                flag = false;
                                break;
                            }
                        }
                    }
                } else {
                    s[i - ] = '';
                }
            }
        }

        for(int i = ; i <= n; i++) {
            if(s[i] == '?') s[i] = '';
        }

        for(int i = ; i <= n; i++) {
            if(is(s[i]) && is(s[i - ])) {
                flag = false;
                break;
            }
        }

        for(int i = ; i < n; i++) {
            if(s[i] == '' && is(s[i - ])) {
                if(s[i + ] >= '' && s[i + ] <= '') {
                    flag = false;
                    break;
                }
            }
        }

        if(is(s[]) || is(s[n])) flag = false;

        if(!flag) puts("IMPOSSIBLE");
        else puts(s + );
    }
    return ;
}
/*
*/

Problem E. Matrix from Arrays

思路：打表找规律发现，n为奇数的时候n * n的矩阵循环，n为偶数时2n * 2n的矩阵循环，然后就二维前缀和搞一搞。

题目没有取模我傻傻地加了取模WA了一次。。

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int N =  + ;
const int mod = 1e9 + ;
LL M[N][N], dp[N][N];
LL A[N];
int n;
LL sum;

void init() {
    int cursor = ;
    for (int i = ; i <= ; ++i) {
        for (int j = ; j <= i; ++j) {
            M[j][i - j] = A[cursor];
            cursor = (cursor + ) % n;
        }
    }
}

LL cal(LL x, LL y) {
    LL cnt1 = x / n, cnt2 = y / n;
    LL ans = sum * cnt1 * cnt2;
    LL num1 = x % n, num2 = y % n;

    ans += dp[num1][num2];
    ans += dp[num1][n - ] * cnt2;
    ans += dp[n - ][num2] * cnt1;
    return ans;
}

int main(){
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d",&n);
        memset(dp, , sizeof(dp));
        memset(M, , sizeof(M));
        sum = ;
        for(int i = ; i < n; i++) scanf("%d", &A[i]);
        init();

        if(n %  == ) n <<= ;

        for(int i = ; i < n; i++) {
            for(int j = ; j < n; j++) {
                sum += M[i][j];
                dp[i][j] += M[i][j];
                if(i) dp[i][j] += dp[i - ][j];
                if(j) dp[i][j] += dp[i][j - ];
                if(i && j) dp[i][j] -= dp[i - ][j - ];
            }
        }

        int q; scanf("%d", &q);
        while(q--) {
            LL x0, y0, x1, y1, ans = ;
            scanf("%lld%lld%lld%lld", &x0, &y0, &x1, &y1);
            ans = cal(x1, y1);
            if(x0 && y0) ans += cal(x0 - , y0 - );
            if(x0) ans -= cal(x0 - , y1);
            if(y0) ans -= cal(x1, y0 - );
            printf("%lld\n", ans);
        }
    }
    return ;
}

/*
*/

Problem D. Nothing is Impossible

思路：这个题无力吐槽，题面改了又改。。 居然还有那么多人用错的题面过了题。。。

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int N =  + ;
const int mod = 1e9 + ;

LL n, m, b[N];
int main(){
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%lld%lld", &n, &m);
        for(int i = ; i <= n; i++) {
            LL a; scanf("%lld%lld", &a, &b[i]);
        }
        LL now = , ans = ;
        sort(b + , b +  + n);
        for(int i = ; i <= n; i++) {
            now *= (b[i] + );
            if(now > m) break;
            ans = i;
        }
        printf("%lld\n", ans);
    }
    return ;
}

/*
*/

Problem J. Let Sudoku Rotate

思路：暴力dfs然后剪枝

#include<bits/stdc++.h>
using namespace std;

bool did[][][];
char s[][];

struct pos{
    int xx,yy,v;
};
vector<pos> v[][][];

int ans;
void Get(int x,int y, int w)
{
    char c;
    if(w <= ) c = w + '';
    else c = 'A' + (w - );

    int xx=x/;
    int yy=y/;

    int i,j;
    bool judge=false;
    for(i=x;i<x+;i++){
        for(j=y;j<y+;j++){
            if(s[i][j]==c){
                judge=true;
                break;
            }
        }
        if(judge) break;
    }
    v[xx][yy][w].push_back({i,j,});

    if(i%==&&j%==){
        v[xx][yy][w].push_back({i,j+,});
        v[xx][yy][w].push_back({i+,j+,});
        v[xx][yy][w].push_back({i+,j,});
    }
    else if(i%==&&j%==){
        v[xx][yy][w].push_back({i+,j,});

        v[xx][yy][w].push_back({i+,j-,});
        v[xx][yy][w].push_back({i,j-,});
    }
    else if(i%==&&j%==){
        v[xx][yy][w].push_back({i-,j,});

        v[xx][yy][w].push_back({i-,j+,});
         v[xx][yy][w].push_back({i,j+,});
    }
    else if(i%==&&j%==){

        v[xx][yy][w].push_back({i,j-,});
        v[xx][yy][w].push_back({i-,j-,});
         v[xx][yy][w].push_back({i-,j,});
    }
    ////

    else if(i%==&&j%==){

        v[xx][yy][w].push_back({i-,j+,});
        v[xx][yy][w].push_back({i+,j+,});
         v[xx][yy][w].push_back({i+,j+,});
    }
    else if(i%==&&j%==){
        v[xx][yy][w].push_back({i-,j-,});

        v[xx][yy][w].push_back({i-,j+,});
         v[xx][yy][w].push_back({i-,j+,});
    }
    else if(i%==&&j%==){
        v[xx][yy][w].push_back({i+,j-,});

        v[xx][yy][w].push_back({i-,j-,});
         v[xx][yy][w].push_back({i-,j-,});
    }
    else if(i%==&&j%==){

        v[xx][yy][w].push_back({i+,j+,});
        v[xx][yy][w].push_back({i+,j-,});
         v[xx][yy][w].push_back({i+,j-,});
    }

    //
    else if(i%==&&j%==){

        v[xx][yy][w].push_back({i-,j+,});
        v[xx][yy][w].push_back({i-,j+,});
         v[xx][yy][w].push_back({i+,j+,});
    }
    else if(i%==&&j%==){
        v[xx][yy][w].push_back({i-,j-,});

        v[xx][yy][w].push_back({i-,j-,});
          v[xx][yy][w].push_back({i-,j+,});
    }
    else if(i%==&&j%==){
        v[xx][yy][w].push_back({i+,j-,});

        v[xx][yy][w].push_back({i+,j-,});
        v[xx][yy][w].push_back({i-,j-,});
    }
    else if(i%==&&j%==){

        v[xx][yy][w].push_back({i+,j+,});
        v[xx][yy][w].push_back({i+,j+,});
        v[xx][yy][w].push_back({i+,j-,});
    }
    ////
    else if(i%==&&j%==){

        v[xx][yy][w].push_back({i,j+,});
        v[xx][yy][w].push_back({i+,j+,});
        v[xx][yy][w].push_back({i+,j,});
    }
    else if(i%==&&j%==){

        v[xx][yy][w].push_back({i+,j,});
        v[xx][yy][w].push_back({i+,j-,});
          v[xx][yy][w].push_back({i,j-,});
    }
    else if(i%==&&j%==){
        v[xx][yy][w].push_back({i-,j,});

        v[xx][yy][w].push_back({i-,j+,});
        v[xx][yy][w].push_back({i,j+,});
    }
    else if(i%==&&j%==){

        v[xx][yy][w].push_back({i,j-,});
        v[xx][yy][w].push_back({i-,j-,});
         v[xx][yy][w].push_back({i-,j,});
    }
}

void dfs(int x,int y,int step){
    if(step>=ans) return ;
    if(x==){
        ans=min(ans,step);
        return ;
    }

        for(int i=;i<;i++)
        {
            bool flag = true;
            for(int w = ; w < ; w++) {
                int tx=v[x][y][w][i].xx,ty=v[x][y][w][i].yy;
                if(did[][tx][w]||did[][ty][w])
                    flag=false;
            }

            //printf("%d %d\n",tx,ty);

            if(flag){
                for(int w=;w<;w++){
                    int tx=v[x][y][w][i].xx,ty=v[x][y][w][i].yy;
                    did[][tx][w]=true;
                    did[][ty][w]=true;
                }
                if(y==){
                    dfs(x+,,step+v[x][y][][i].v);
                }
                else{
                    dfs(x,y+,step+v[x][y][][i].v);
                }
                for(int w=;w<;w++){
                    int tx=v[x][y][w][i].xx,ty=v[x][y][w][i].yy;
                    did[][tx][w]=false;
                    did[][ty][w]=false;
                }

            }

        }

}

int main(){
    int T; scanf("%d",&T);
    while(T--){
        memset(did,false,sizeof(did));
        for(int i=;i<;i++)
            scanf("%s",s[i]);

        for(int i=;i<;i++)
            for(int j=;j<;j++)
                for(int w = ; w < ; w++)
                    v[i][j][w].clear();

        for(int i=;i<;i+=)
        {
            for(int j=;j<;j+=) {
                for(int k = ; k < ; k++)
                    Get(i, j, k);
            }
        }

        ans=;
        dfs(,,);
        printf("%d\n",ans);
    }
}

补题****************************************************************

Problem B. Harvest of Apples

思路：这个莫队真的不好想，首先得推出公式来。。。

S(n, m + 1) = S(n, m) + C(n, m + 1)

S(n + 1, m) = 2 * S(n, m) - C(n, m)

然后搞莫队。。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int,int>
#define pLL pair<long long, long long>
#define piii pair<int, pair<int,int>>

using namespace std;

const int N = 1e5 + ;
const int M = 1e4 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;
const double PI = acos(-);
const int B = ;

int ans[N], f[N], inv[N];
LL ret;
struct Qus {
    int n, m, id;
    bool operator < (const Qus &rhs) const {
        if(n / B == rhs.n / B) return m < rhs.m;
        return n / B < rhs.n / B;
    }
} qus[N];

int fastPow(int a, int b) {
    int ans = ;
    while(b) {
        if(b & ) ans = 1ll * ans * a % mod;
        a = 1ll * a * a % mod; b >>= ;
    }
    return ans;
}

void init() {
    f[] = ;
    for(int i = ; i < N; i++) f[i] = 1ll * f[i - ] * i % mod;
    inv[N - ] = fastPow(f[N - ], mod - );
    for(int i = N - ; i >= ; i--) inv[i] = 1ll * inv[i + ] * (i + ) % mod;
}

int comb(int n, int m) {
    return 1ll * f[n] * inv[m] % mod * inv[n - m] % mod;
}

void update1(int m, int n, int op) {
    op = -op;
    ret += op * comb(n, m + );
    if(ret >= mod) ret -= mod;
    if(ret < ) ret += mod;
}

void update2(int m, int n, int op) {
    if(op == ) {
        ret += ret; if(ret >= mod) ret -= mod;
        ret -= comb(n - , m); if(ret < ) ret += mod;
    } else {
        ret += comb(n - , m); if(ret >= mod) ret -= mod;
        ret = ret * inv[] % mod;
    }
}

int main() {
    init();
    int q; scanf("%d", &q);
    for(int i = ; i <= q; i++) {
        scanf("%d%d", &qus[i].n, &qus[i].m);
        qus[i].id = i;
    }

    sort(qus + , qus + q + );

    ret = ;
    int l = , r = ;
    for(int i = ; i <= q; i++) {
        int L = qus[i].m, R = qus[i].n;
        while(r < R) update2(l, ++r, );
        while(l > L) update1(--l, r, );
        while(r > R) update2(l, r--, -);
        while(l < L) update1(l++, r, -);
        ans[qus[i].id] = ret;
    }
    for(int i = ; i <= q; i++) printf("%d\n", ans[i]);
    return ;
}

/*
*/