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Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16592 Accepted Submission(s):
5656
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 int dp[101][7500]; 6 int main() 7 { 8 int coin[5]= {1,5,10,25,50}; 9 int i,j,n,m,k; 10 memset(dp,0,sizeof(dp)); 11 dp[0][0]=1; 12 for(i=0; i<5; i++) //用的硬币 13 { 14 for(k=1; k<=100; k++) //硬币数目 15 { 16 for(j=coin[i]; j<=7500; j++) 17 { 18 dp[k][j]+=dp[k-1][j-coin[i]]; 19 } 20 } 21 22 } 23 while(~scanf("%d",&n)) 24 { 25 int ans=0; 26 for(i=0; i<=100; i++) 27 ans+=dp[i][n]; 28 printf("%d\n",ans); 29 } 30 return 0; 31 }
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原文地址:http://www.cnblogs.com/pshw/p/5171223.html