hdu 2069 二维母函数

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

题意：

能不能用1 5 10 25 50 组成输入的数，并且使用的个数不能超过100个

题解：

多加一维维护个数。c1[i][j] 表示 c1[结果][组成该结果的个数]。
预处理优化一下。

代码：

#include <bits/stdc++.h>using namespace std;int c1[251][101];int c2[251][101];//次数顶多为250 前一个括号是指系数 后一个指硬币个数int val[6]={0,1,5,10,25,50};int main(){    memset(c1,0,sizeof(c1));    memset(c2,0,sizeof(c2));    int i,j,k,n,l,sum;    for(i=0;i<=100;i++)    {        c1[i][i]=1;    }    n=250;    for(i=2;i<=5;i++)    {        for(j=0;j<=n;j++)            for(k=0;k+j<=n;k+=val[i])             for(l=0;l+k/val[i]<=100;l++)               c2[k+j][l+k/val[i]]+=c1[j][l];        for(k=0;k<=n;k++)            for(l=0;l<=100;l++)        {            c1[k][l]=c2[k][l];            c2[k][l]=0;        }    }    while(cin>>n)    {        sum=0;        for(i=0;i<=100;i++)            sum+=c1[n][i];        cout<<sum<<endl;    }    return 0;}