UVA674- Coin Change

来源：互联网发布：什么是资本化知乎编辑：程序博客网时间：2024/05/18 03:37

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;}

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