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;

}