UVA 10912 Simple Minded Hashing

题意就略了。刚一看被数据吓住了。看到字符要求严格递增。那么如果字串长大于26那必然方案数目为0；同时1+2+3....+24+25+26=351如果大于这个数也是不可能的

令dp[i][j][k]表示第i位为第j个字符和为K时的方案数目

那么 dp[i][j][k]=sum(dp[i-1][m][k-t])  {m<j;k-t尝试将第j位置为t}

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}
#define MAXN 360
#define MAXD 30
int l,s;
int tab[MAXD];
int dp[MAXD][MAXD][MAXN];
void init()
{
    memset(dp,0,sizeof(dp));
    memset(tab,0,sizeof(tab));
    for (int i=1; i<=26; i++) tab[i] = tab[i-1] + i;
    for (int i=1; i<=26; i++) dp[1][i][i] = 1;
    for (int i=2; i<=26; i++)
        for (int j=i; j<=26; j++)
            for (int s=tab[i]; s<=351; s++)
    {
        for (int k=1; k<j && k<s; k++)
            dp[i][j][s] += dp[i-1][k][s-j];
    }
}
int slove()
{
    int ans = 0;
    for (int i=l; i<=26; i++)
        ans += dp[l][i][s];
    return ans;
}
int main()
{
    init();
    int kase = 1;
    while (scanf("%d%d",&l,&s)!=EOF)
    {
        if (l == 0 && s==0) break;
        if (l > 26 || s > 351)
        printf("Case %d: %d
",kase++,0);
        else printf("Case %d: %d
",kase++,slove());
    }
    return 0;
}