Coin Change
Time Limit: 1000ms
Memory Limit: 32768KB
This problem will be judged on HDU. Original ID: 206964-bit integer IO format: %I64d Java class name: Main
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.
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：此题限制了银币数，故拓展了一维。。dp[i][j]表示使用了i枚硬币获取j的价值
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 8000; 4 int n,dp[102][maxn],cent[] = {1,5,10,25,50}; 5 int main(){ 6 while(~scanf("%d",&n)){ 7 memset(dp,0,sizeof dp); 8 dp[0][0] = 1; 9 for(int i = 0; i < 5; ++i) 10 for(int j = 1; j <= 100; ++j) 11 for(int k = cent[i]; k <= n; ++k) 12 dp[j][k] += dp[j-1][k-cent[i]]; 13 int ans = 0; 14 for(int i = 0; i <= 100; ++i) 15 ans += dp[i][n]; 16 printf("%d ",ans); 17 } 18 return 0; 19 }