hdu 1398(普通型母函数)

最新推荐文章于 2019-09-30 17:48:40 发布

原创最新推荐文章于 2019-09-30 17:48:40 发布 · 351 阅读

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#combinations #integer #shapes #numbers #input #each

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本文探讨了一种使用特定面额的方币(SquareCoins)进行支付的问题，这些方币的面额为平方数。文章提供了一个算法示例，演示如何计算用这些特殊硬币支付给定金额的不同组合的数量。

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Square Coins

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4675 Accepted Submission(s): 3166

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

Sample Output

Source

Asia 1999, Kyoto (Japan)

Recommend

Ignatius.L

题目类型：普通型母函数

题目描述：略

题目分析：裸

代码如下：

#include <stdio.h>
#include <memory.h>
#define N 301

int r[N];
int t[N];

void init(){
    int i,j,k;
    for( i = 0; i < N; i++){
        r[i] = 1;
        t[i] = 0;
    }
    for( k = 2; k <= 17; k++){
        for( i = 0; i < N; i++){
            for( j = 0; j < N; j += k*k) {
                if( r[i] != 0 && i + j < N){
                    t[i+j] += r[i];
                }
            }
        }
        for( i = 0; i < N ; i++){
            r[i] = t[i];
            t[i] = 0;
        }
    }
}



int main()
{
    int x;
    init();
    while(scanf("%d",&x),x){
        printf("%d\n",r[x]);
    }

    return 0;
}