UVA 10912 Simple Minded Hashing

题意：求长度为L状态为S的方法有多少，用三维表示前i个字符组成长度为j的状态是S的个数有多少，看了题解才知道用三维的，开始想到的是二维的，功力不够啊，想到的话就不会难了，

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int f[27][27][355];int main(){    memset(f,0,sizeof(f));    f[0][0][0] = 1;    int sum = 355;    for (int i = 1; i <= 26; i++)        for (int j = 0; j <= i; j++)            for (int k = 0; k <= 351; k++){                f[i][j][k] = f[i-1][j][k];                if (j && k >= i)                    f[i][j][k] += f[i-1][j-1][k-i];            }                    int cas = 1,l,s;    while (scanf("%d%d",&l,&s) != EOF && l+s){        int ans = (l>26||l<1||s>351||s<1)?0:f[26][l][s];        printf("Case %d: %d\n",cas++,ans);    }        return 0;}