Uva 674 Coin Change
2012-10-08 09:04
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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, two 5-cent coins and one 1-cent coin, 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 7489 cents.
Input
The input file contains any number of lines, each one consisting of a number 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
完全背包求解方案数问题
F[i]表示可达且可达的方案数!
状态转移方程:f[i]=f[i]+f[i-c[k]];
For example, if we have 11 cents, then we can make changes with one 10-cent coin and one 1-cent coin, two 5-cent coins and one 1-cent coin, 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 7489 cents.
Input
The input file contains any number of lines, each one consisting of a number 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
完全背包求解方案数问题
F[i]表示可达且可达的方案数!
状态转移方程:f[i]=f[i]+f[i-c[k]];
#include<stdio.h> #include<string.h> int f[8000]; int c[5]={1,5,10,25,50}; int main(void) { int v,k,i; f[0]=1; for(k=0;k<5;k++) for(i=1;i<=7489;i++) { if(i>=c[k]) //这里不能忽略,不然指针访问出问题 f[i]+=f[i-c[k]]; } while(scanf("%d",&v)==1) { printf("%d\n",f[v]); } return 0; }
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