Coin Change(hdu2069,暴力or母函数)
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http://acm.hdu.edu.cn/showproblem.php?pid=2069
Coin Change
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10585 Accepted Submission(s): 3532
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
Author
Lily
Source
浙江工业大学网络选拔赛
Recommend
linle
解析:
由于数据比较小。所以打表是可行的
*/
#include<stdio.h>#include<string.h>#include<math.h>//#include<stdlib.h>#include<algorithm>#include<iostream>const int maxn=410;int b[6]={1,5,10,25,50};int main(){int i,j,k,g,f;int n;while(scanf("%d",&n)!=EOF){int ans=0,t,m=100;for(i=0;i<=n;i++) { for(j=0;j<=(n-i*b[0])/b[1];j++) { for(k=0;k<=(n-i*b[0]-j*b[1])/b[2];k++) { for(g=0;g<=(n-i*b[0]-j*b[1]-k*b[2])/b[3];g++){ for(f=0;f<=(n-i*b[0]-j*b[1]-k*b[2]-g*b[3])/b[4];f++) { if(n-i*b[0]-j*b[1]-k*b[2]-g*b[3]-f*b[4]==0&&(i+j+k+g+f)<=m) ans++; } } } }} printf("%d\n",ans);}return 0;}还有一种做法:母函数法:看了标程的:#include<stdio.h>#include<string.h>#include <iostream>using namespace std;int c2[255][105],c1[255][105];//c[i][j],表示凑成i需要j个数的组合数int a[6]={1,5,10,25,50};void init(){int i,j,t,k;c1[0][0]=1;for(i=0;i<5;i++) { for(j=0;j<=250;j++) { for(k=-0;k*a[i]+j<=250;k++) { for(t=0;k+t<=100;t++) c2[j+k*a[i]][k+t]+=c1[j][t]; } } memcpy(c1,c2,sizeof(c2)); memset(c2,0,sizeof(c2)); }}int main(){ init(); int n; while(scanf("%d",&n)!=EOF) { int ans=0; for(int i=0;i<=100;i++) ans+=c1[n][i]; printf("%d\n",ans); } return 0;}
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