杭电1398 Square Coins 简单母函数

来源:互联网 发布:c语言 字符串统计函数 编辑:程序博客网 时间:2024/06/11 17:35

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10557    Accepted Submission(s): 7210


Problem Description
People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values of all square numbers up to 289 (=17^2), i.e., 1-credit coins, 4-credit coins, 9-credit coins, ..., and 289-credit coins, are available in Silverland. 
There are four combinations of coins to pay ten credits: 

ten 1-credit coins,
one 4-credit coin and six 1-credit coins,
two 4-credit coins and two 1-credit coins, and
one 9-credit coin and one 1-credit coin. 

Your mission is to count the number of ways to pay a given amount using coins of Silverland.
 

Input
The input consists of lines each containing an integer meaning an amount to be paid, followed by a line containing a zero. You may assume that all the amounts are positive and less than 300.
 

Output
For each of the given amount, one line containing a single integer representing the number of combinations of coins should be output. No other characters should appear in the output. 
 

Sample Input
210300
 

Sample Output
1427
 

Source
Asia 1999, Kyoto (Japan)
 

Recommend
Ignatius.L   |   We have carefully selected several similar problems for you:  2152 2082 1709 2079 2110 
 


简单母函数:

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int a[11000],a1[11000],ans[11000],i,j,k,l,m,n,p;int main(){while(scanf("%d",&n),n){for(i=0;i<=n;i++){a[i]=1;a1[i]=0;}for(i=2;i<=17;i++){for(j=0;j<=n;j++)for(k=0;k+j<=n;k+=i*i)a1[j+k]+=a[j];for(k=0;k<=n;k++){a[k]=a1[k];a1[k]=0;}}printf("%d\n",a[n]);}}


0 0