Coin Change

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 19239    Accepted Submission(s): 6719 Problem Description Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We want to make changes with these coins for a given amount of money. For example, if we have 11 cents, then we can make changes with one 10-cent coin and one 1-cent coin, or two 5-cent coins and one 1-cent coin, or one 5-cent coin and six 1-cent coins, or eleven 1-cent coins. So there are four ways of making changes for 11 cents with the above coins. Note that we count that there is one way of making change for zero cent. Write a program to find the total number of different ways of making changes for any amount of money in cents. Your program should be able to handle up to 100 coins.   Input The input file contains any number of lines, each one consisting of a number ( ≤250 ) for the amount of money in cents.   Output For each input line, output a line containing the number of different ways of making changes with the above 5 types of coins.   Sample Input 11 26   Sample Output 4 13

解题报告：暴力求解

code：

#include<cstdio> #include<iostream> #include<cstring> #include<string> #include<sstream> #include<algorithm> #include<math.h> #include<queue> #include<stack> #include<map> #include<set> using namespace std; typedef long long ll; const int maxn=255; int main() { // freopen("input.txt","r",stdin); int n; while(~scanf("%d",&n)){ int sum=0; for(int i=0;i*50<=n;i++) for(int j=0;j*25<=n;j++) for(int k=0;k*10<=n;k++) for(int p=0;p*5<=n;p++)/*别忘了最多只能有100个硬币*/ if((i*50+j*25+k*10+p*5)<=n && (i+j+k+p+(n-(i*50+j*25+k*10+p*5)))<=100) sum++; printf("%d\n",sum); } return 0; }