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HDU 1398 Square Coins(母函数)
Square Coins
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
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.
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
#include<stdio.h>#define MAX 310int main(void){ int n,i,j,k; int c1[MAX],c2[MAX]; while(scanf("%d",&n)!=EOF&&n) { for(i=0;i<=n;i++) { c1[i]=1; c2[i]=0;//当前方程 } for(i=2;i*i<=n;i++){ for(j=0;j<=n;j++){ for(k=0;k+j<=n;k+=i*i) c2[k+j]+=c1[j];//当前方程的系数 向下继承k(各种硬币直)+j } for(j=0;j<=n;j++){ c1[j]=c2[j]; c2[j]=0; } } printf("%d\n",c1[n]); } return 0;}
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