Codeforces 754A(搜索)

设s[i][j]为序列i到j的和，当s[i][j]≠0时，即可从i跳到j+1.目标为从1跳到n+1，所以按照题意暴力即可。

#include <bits/stdc++.h>using namespace std;#define rep(i,a,b)              for(int i(a); i <= (b); ++i)
#define dec(i,a,b)              for(int i(a); i >= (b); --i)const int Q     =    1000        +       10;struct node{int x, y;
} ans[Q];
int s[Q][Q];
int a[Q];bool flag = false;
int n;
int sum;void print(int x){printf("YES\n%d\n", x);rep(i, 1, x) printf("%d %d\n", ans[i].x, ans[i].y);
}void dfs(int x, int step){if (flag) return;if (x == n + 1){flag = true;print(step - 1);return;}dec(i, n, x) if (s[x][i]){ans[step].x = x, ans[step].y = i;dfs(i + 1, step + 1);}
}   int main(){scanf("%d", &n);rep(i, 1, n) scanf("%d", a + i);rep(i, 1, n){sum = 0;rep(j, i, n){sum += a[j];s[i][j] = sum;}}dfs(1, 1);if (!flag) puts("NO");return 0;}

转载于:https://www.cnblogs.com/cxhscst2/p/6648798.html

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Codeforces 754A(搜索)

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