题目链接
In a strange shop there are n types of coins of value A1, A2 ... An. C1, C2, ... Cn denote the number of coins of value A1, A2 ... An 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。
代码语言：javascript复制#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: %dn", 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
代码语言：javascript复制#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: %dn", tc, dp[k]);
}
return 0;
}第三题又是完全背包
代码语言：javascript复制#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: %dn", ca, ans);
}
return 0;
}
