HDU 2069 Coin Change

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

1126

 

Sample Output

413

 

//枚举#include <stdio.h>#include <string.h>#define maxn 251int main(){int i, j, k, m, n,sum,t;int d[maxn],count[maxn];memset(count, 0, sizeof(count));count[0] = 1;for (i = 1;i < maxn;i++) {d[i] = i;//d[i]代表i美分for (j = 0;j <= d[i] / 50;j++) {for (k = 0;k <= d[i] / 25;k++) {for (m = 0;m <= d[i] / 10;m++) {for (n = 0;n <= d[i] / 5;n++) {sum = 50 * j + 25 * k + 10 * m + 5 * n;t=j + k + m + n;if (sum <= d[i] && (t + d[i] - sum <= 100)) count[i]++;}}}}}int x;while (~scanf("%d", &x)) {printf("%d\n", count[x]);}return 0;}

//母函数#include <stdio.h>#include <string.h>#define maxn 251int main(){int c1[maxn][110], c2[maxn][110], i, j, k, l, c[maxn];//c1,c2多开一维是为了判别总的硬币数量是否超过100//c1[i][j]表示i美分时需要j枚硬币，c[i]表示i美分时方案数int a[6] = { 0,1,5,10,25,50 }, b[6];//数组a保存可选硬币种类,数组b保存每种硬币最多可选个数memset(c1, 0, sizeof(c1));memset(c2, 0, sizeof(c2));memset(c, 0, sizeof(c));c1[0][0] = 1;for (i = 1;i <= 5;i++) {for (j = 0;j <= 250;j++)for (k = 0;k*a[i] + j <= 250;k++)//新加入的硬币个数k个for (l = 0;l + k <= 100;l++)//l是原先已用硬币数c2[k*a[i] + j][l+k] += c1[j][l];//有可能出现c1[j][l]为0，此时l个硬币得不到j美分memcpy(c1, c2, sizeof(c1));memset(c2, 0, sizeof(c2));}for (i = 1;i <= 250;i++)for (j = 0;j <= 100;j++)c[i] += c1[i][j];//将满足凑成i美分时所用硬币数<=100的方案数累加c[0] = 1;//题目要求0美分时是一种方案int n;while (~scanf("%d", &n)) {printf("%d\n", c[n]);}return 0;}