Uva 11137 Ingenuous Cubrency(DP)

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=23&page=show_problem&problem=2078

Ingenuous Cubrency

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

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Description

Problem I: Ingenuous Cubrency

People in Cubeland use cubic coins. Not only the unit of currency is called a cube but also the coins are shaped like cubes and their values are cubes. Coins with values of all cubic numbers up to 9261 (= 21 ³), i.e., coins with the denominations of 1, 8, 27, ..., up to 9261 cubes, are available in Cubeland.

Your task is to count the number of ways to pay a given amount using cubic coins of Cubeland. For example, there are 3 ways to pay 21 cubes: twenty one 1 cube coins, or one 8 cube coin and thirteen 1 cube coins, or two 8 cube coin and five 1 cube coins.

Input consists of lines each containing an integer amount to be paid. You may assume that all the amounts are positive and less than 10000.

For each of the given amounts to be paid output one line containing a single integer representing the number of ways to pay the given amount using the coins available in Cubeland.

Sample input

10 21779999

Output for sample input

2322440022018293

题目意思：求一个数可以用几个数的立方和表示，有几种表示方法

#include <stdio.h>#include <stdlib.h>#include <string.h>long long dp[10001];long long c[25];int main(){    memset(dp,0,sizeof(dp));    int i,j;    for(i=1Q;i<=21;i++)        c[i]=i*i*i;    dp[0]=1; //当当前数字等于另外一个数的立方时，算一种解    for(i=1;i<=21;i++) //每一个立方数对dp数组产生的影响    {        for(j=c[i];j<=10000;j++)        {            dp[j] += dp[j-c[i]];//利用以前求出的结果        }    }    int a;    while(scanf("%d",&a)!=EOF)    {        printf("%lld\n",dp[a]);    }    return 0;}