独木桥

怎么说呢

#include<iostream>

#include<cstdio>

#include<algorithm>

using namespace std;

const int N = ;

int n, l, pos[N], maxx = , minn = ;

int main(){

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

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

    sort(pos + , pos + n + );

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

        if(pos[i] <= ( + l) / ) maxx = max(maxx, l - pos[i] + ), minn = max(pos[i], minn);

        else maxx = max(maxx, pos[i]), minn = max(minn, l - pos[i] + );

    printf("%d %d", minn, maxx);

    return ;

}

传纸条 & 方格取数

多维dp的应用

code传纸条

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<algorithm>

#include<cmath>

using namespace std;

const int N = ;

int n, m, a[N][N], dp[N][N][N][N];

inline int myMax(int u, int v, int x, int y){

    int ret = u;

    ret = max(ret, v);

    ret = max(ret, x);

    ret = max(ret, y);

    return ret;

}

int main(){

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

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

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

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

    dp[][][][] = a[][];

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

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

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

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

                    if(i ==  && j == ) continue;

                    dp[i][j][k][l] = myMax(dp[i - ][j][k - ][l], dp[i][j - ][k - ][l],

                                           dp[i - ][j][k][l - ], dp[i][j - ][k][l - ])

                                     + a[i][j] + a[k][l];

                    if(i == k && j == l)

                        dp[i][j][k][l] -= a[i][j];

                }

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

    return ;

}

code方格取数

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<algorithm>

#include<cmath>

using namespace std;

const int N = ;

int n, a[N][N], dp[N][N][N][N];

inline int myMax(int u, int v, int x, int y){

    int ret = u;

    ret = max(ret, v);

    ret = max(ret, x);

    ret = max(ret, y);

    return ret;

}

int main(){

    scanf("%d", &n);

    int x, y, w;

    while(scanf("%d%d%d", &x, &y, &w), x + y + w)

        a[x][y] = w;

//    dp[1][1][1][1] = a[1][1];

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

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

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

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

//                    if(i == k && j == l && i != n && j != n) continue;

                    dp[i][j][k][l] = myMax(dp[i - ][j][k - ][l], dp[i][j - ][k - ][l],

                                           dp[i - ][j][k][l - ], dp[i][j - ][k][l - ])

                                     + a[i][j] + a[k][l];

                    if(i == k && j == l )

                        dp[i][j][k][l] -= a[i][j];

                }

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

    return ;

}

矩阵取数游戏

问题识破 + dp + 高精度

#include<iostream>

#include<cstdio>

#include<string>

#include<cstring>

using namespace std;

inline int read(){

    int i = , f = ; char ch = getchar();

    for(; (ch < '' || ch > '') && ch != '-'; ch = getchar());

    if(ch == '-') f = -, ch = getchar();

    for(; ch >= '' && ch <= ''; ch = getchar())

        i = (i << ) + (i << ) + (ch - '');

    return i * f;

}

const int N = ;

int n, m;

struct bign{

    int len, s[N];

    bign():len(){memset(s, , sizeof s);}

    inline void clear(){

        while(len >  && s[len] == ) len--;

    }

    inline bign operator * (const bign &u) const{

        bign ret;

        ret.len = len + u.len + ;

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

            for(int j = ; j <= u.len; j++)

                ret.s[i + j - ] += s[i] * u.s[j];

        for(int i = ; i <= ret.len; i++)

            if(ret.s[i] >= ){

                ret.s[i + ] += ret.s[i] / ;

                ret.s[i] %= ;

            }

        ret.clear();

        return ret;

    }

    inline void print(){

        clear();

        for(int i = len; i >= ; i--)

            putchar(s[i] + '');

    }

    inline bool operator > (const bign &u) const{

        if(len != u.len) return len > u.len;

        for(int i = len; i >= ; i--)

            if(s[i] != u.s[i]) return s[i] > u.s[i];

        return false;

    }

    inline bign operator + (const bign &u) const{

        bign ret;

        ret.len = ;

        int i, g;

        for(i = , g = ; g || i <= max(len, u.len); i++){

            ret.s[++ret.len] = g;

            if(i <= len) ret.s[ret.len] += s[i];

            if(i <= u.len) ret.s[ret.len] += u.s[i];

            g = ret.s[ret.len] / ;

            ret.s[ret.len] %= ;

        }

        ret.clear();

        return ret;

    }

}pow2[N][N], f[N][N], big2, ans, ret, a[N], big0;

inline bign newBign(int x){

    bign ret;

    ret.len = ;

    while(x){

        ret.s[++ret.len] = x % ;

        x /= ;

    }

    if(ret.len == ) ret.len = ;

    return ret;

}

inline void initPow(){

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

        for(int j = ; j <= m; j++)

            pow2[i][j] = pow2[i][j - ] + pow2[i][j - ];

}

inline bign max(bign a, bign b){

    return a > b ? a : b;

}

int main(){

    n = read(), m = read();

    ans = big0 = newBign(); big2 = newBign();

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

        ret = big0;

        for(int i = ; i <= m; i++) pow2[i][] = newBign(read());

        initPow();

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

            for(int j = ; j <= m; j++)

                f[i][j] = big0;

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

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

                if(i > ) f[i][j] = max(f[i - ][j] + pow2[i][i + j], f[i][j]);

                if(j > ) f[i][j] = max(f[i][j - ] + pow2[m - j + ][i + j], f[i][j]);

            }

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

            ret = max(ret, f[i][m - i]);

        ans = ans + ret;

    }

    ans.print();

    return ;

}

铺地毯

怎么说呢

#include<iostream>

#include<cstdio>

using namespace std;

const int N = 1e4 + ;

int n, x1[N], y1[N], l[N], w[N], x, y, ans = -;

int main(){

    scanf("%d", &n);

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

        scanf("%d%d%d%d", &x1[i], &y1[i], &l[i], &w[i]);

    scanf("%d%d", &x, &y);

    for(int i = n; i >= ; i--){

        if(x1[i] <= x && x <= x1[i] + l[i] -  && y1[i] <= y && y <= y1[i] + w[i] - ){

            ans = i;

            break;

        }

    }

    printf("%d", ans);

    return ;

}

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