题意：用所给的硬币面值构成所需的面值

思路：因为所用硬币数量不限，所以很容易想到完全背包。

递推：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN = 10005;
int n;

int coin[] = {1, 5, 10, 25, 50};
long long d[MAXN];

void dp() {
    memset(d, 0, sizeof(d));
    d[0] = 1;
    for (int i = 0; i < 5; i++)
        for (int j = coin[i]; j < MAXN; j++)
            d[j] += d[j - coin[i]];
}

int main() {
    dp();
    while (scanf("%d", &n) != EOF) {
        printf("%lld\n", d[n]);          
    } 
    return 0;
}

记忆化搜索：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN = 10005;
int n;

int coin[] = {1, 5, 10, 25, 50};
long long d[MAXN][6];

void init() {
    memset(d, -1, sizeof(d));
    for (int i = 0; i < 5; i++)
        d[0][i] = 1;
}

long long dp(int s, int m) {
    if (d[s][m] != -1)
        return d[s][m];
    d[s][m] = 0;
    for (int i = m; i < 5 && s >= coin[i]; i++)
        d[s][m] += dp(s - coin[i], i);
    return d[s][m];
}

int main() {
    init();
    while (scanf("%d", &n) != EOF) {
        long long ans = dp(n, 0); 
        printf("%lld\n", ans);
    } 
    return 0;
}