uva 10313

递推   参考了别人的解法 

dp[i][j] 表示价值为i用j个硬币可以有多少种方法   

dp[j][k] += dp[j-i][k-1] 意思是多加一枚价值为i的硬币，加上价值为j-i用k-1个硬币的总数

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 305
#define INF 0x7fffffff
#define inf 10000000
#define ull unsigned long long
#define ll long long
using namespace std;

ll dp[maxn][maxn];
char s[maxn];

void init()
{
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for(int i = 1; i < maxn; ++ i)
        for(int j = i; j < maxn; ++ j)
            for(int k = 1; k < maxn; ++ k)
                dp[j][k] += dp[j-i][k-1];
}

int main()
{
    init();
    int n, a, b;
    while(gets(s) != NULL)
    {
        ll ans = 0;
        int flag = sscanf(s, "%d%d%d", &n, &a, &b);
        a = min(a, 300);
        b = min(b, 300);
        if(flag == 1)
            for(int i = 0; i <= n; ++ i)
                ans += dp[n][i];
        else if(flag == 2)
            for(int i = 0; i <= a; ++ i)
                ans += dp[n][i];
        else
            for(int i = a; i <= b; ++ i)
                ans += dp[n][i];
        printf("%lld\n", ans);
    }
    return 0;
}