light oj 12311232 1233 Coin Change 背包

题目链接

In a strange shop there are n types of coins of value A₁, A₂ ... A_n. C₁, C₂, ... C_n denote the number of coins of value A₁, A₂ ... A_n respectively. You have to find the number of ways you can make K using the coins.

For example, suppose there are three coins 1, 2, 5 and we can use coin 1 at most 3 times, coin 2 at most 2 times and coin 5 at most 1 time. Then if K = 5 the possible ways are:

1112

122

5  ………………………………

暂时到半懂不懂也没办法讲明白，就不误人子弟了，直接贴代码了。

#include <stdio.h>
#include <string.h>
const int mod = 100000007;
int dp[1005];
int coin[55];
int cnt[55];

int main()
{
    int t, n, k;
    scanf("%d", &t);
    for (int tc = 1; tc <= t; tc++)
    {
        scanf("%d %d", &n, &k);
        for (int i = 1; i <= n; i++)
        {
            scanf("%d", &coin[i]);
        }
        for (int j = 1; j <= n; j++)
        {
            scanf("%d", &cnt[j]);
        }
        memset(dp, 0, sizeof(dp));
        dp[0] = 1;
        for (int i = 1; i <= n; i++)
        {
            for (int j = k; j >= 0; j--)
            {
                for (int l = 1; l <= cnt[i]; l++)
                {
                    if (j - l*coin[i] >= 0)
                    dp[j] += dp[j-coin[i]*l];
                }
            }
            for (int j = 0; j <= k; j++)
                dp[j] %= mod;
        }
        printf("Case %d: %d\n", tc, dp[k]);
    }
    return 0;
}

如果说把第一题看做01背包的话，这一题就是完全背包了 

#include <stdio.h>
#include <string.h>
const int mod = 100000007;
int dp[10050];
int coin[105];

int main()
{
    int t, n, k;
    scanf("%d", &t);
    for (int tc = 1; tc <= t; tc++)
    {
        scanf("%d %d", &n, &k);
        for (int i = 1; i <= n; i++)
        {
            scanf("%d", &coin[i]);
        }

        memset(dp, 0, sizeof(dp));
        dp[0] = 1;
        for (int i = 1; i <= n; i++)
        {
            for (int j = coin[i]; j <= k; j++)
            {
                dp[j] += dp[j-coin[i]];
                dp[j] %= mod;
            }
        }
        printf("Case %d: %d\n", tc, dp[k]);
    }
    return 0;
}

第三题又是完全背包

#include <stdio.h>
#include <string.h>

int dp[100005];
int coin[101];
int cnt[101];
int used[1000101];

int main()
{
    int t, n, k;
    scanf("%d", &t);
    for (int ca = 1; ca <= t; ca++)
    {
        scanf("%d %d", &n, &k);
        for (int i = 1; i <= n; i++)
        {
            scanf("%d", &coin[i]);
        }
        for (int j = 1; j <= n; j++)
        {
            scanf("%d", &cnt[j]);
        }
        memset(dp, 0, sizeof(dp));
        dp[0] = 1;
        int ans = 0;
        for (int i = 1; i <= n; i++)
        {
            memset(used, 0, sizeof(used));
            for (int j = coin[i]; j <= k; j++)
            {
                if (!dp[j] && dp[j-coin[i]] && used[j-coin[i]] < cnt[i])
                {
                    ans++;
                    used[j]=used[j-coin[i]]+1;
                    dp[j] = 1;
                }
            }
        }
        printf("Case %d: %d\n", ca, ans);
    }
    return 0;
}