HDU 2448 Coins dp
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CoinsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13732 Accepted Submission(s): 5492
Problem Description
Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse and found there were some coins. He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.
You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.
Input
The input contains several test cases. The first line of each test case contains two integers n(1 ≤ n ≤ 100),m(m ≤ 100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1 ≤ Ai ≤ 100000,1 ≤ Ci ≤ 1000). The last test case is followed by two zeros.
Output
For each test case output the answer on a single line.
Sample Input
3 10 1 2 4 2 1 1 2 5 1 4 2 1 0 0
Sample Output
8 4
Source
2009 Multi-University Training Contest 3 - Host by WHU
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题目大意：给定若干面值的硬币，每种面值的硬币有一定数目，问这些硬币可以组成多少不超过m面值的数额。
解题思路：dp[i][j]为前I种硬币组成面值j，第i种硬币所剩个数。dp[i][j]=-1代表组成不了j面值。
当dp[i-1][j]>=0时，dp[i][j] = c[i];
当j < a[i],或者dp[i][j-a[]]>=0时，dp[i][j] = -1;
其他情况dp[i][j] = dp[i - 1][j - a[i]] - 1;
注：用此种方法可防止TLE,最后利用循环数组防止MLE。
#include
#include
#include
#include
using namespace std;
const int maxn = 105;
const int maxm = 1e5 + 100;
int a[105], c[105];
int dp[maxm];
int main()
{
int n, m;
while(scanf("%d%d", &n, &m) != EOF) {
if(n == 0 && m == 0) {
break;
}
for(int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
}
memset(dp, -1, sizeof(dp));
for(int i = 1; i <= n; ++i) {
scanf("%d", &c[i]);
}
dp[0] = 0;
for(int i = 1; i <= n; ++i) {
for(int j = 0; j <= m; ++j) {
if(dp[j] >= 0) {
dp[j] = c[i];
} else if(j < a[i] || dp[j - a[i]] <= 0) {
dp[j] = -1;
} else {
dp[j] = dp[j - a[i]] - 1;
}
}
}
int ans = 0;
for(int i = 1; i <= m; ++i) {
if(dp[i] >= 0) {
ans++;
}
}
printf("%d\n", ans);
}
return 0;
}