您的位置：首页 > Web前端

POJ 1543 Perfect Cubes

2016-06-12 19:30 197 查看

Language:DefaultPerfect Cubes

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15362		Accepted: 7956

DescriptionFor hundreds of years Fermat's Last Theorem, which stated simply that for n > 2 there exist no integers a, b, c > 1 such that a^n = b^n + c^n, has remained elusively unproven. (A recent proof is believed to be correct, though it is still undergoing scrutiny.)It is possible, however, to find integers greater than 1 that satisfy the "perfect cube" equation a^3 = b^3 + c^3 + d^3 (e.g. a quick calculation will show that the equation 12^3 = 6^3 + 8^3 + 10^3 is indeed true). This problem requires that you write a programto find all sets of numbers {a,b,c,d} which satisfy this equation for a <= N.InputOne integer N (N <= 100).OutputThe output should be listed as shown below, one perfect cube per line, in non-decreasing order of a (i.e. the lines should be sorted by their a values). The values of b, c, and d should also be listed in non-decreasing order on the line itself. There do existseveral values of a which can be produced from multiple distinct sets of b, c, and d triples. In these cases, the triples with the smaller b values should be listed first.Sample Input

Sample Output

Cube = 6, Triple = (3,4,5)
Cube = 12, Triple = (6,8,10)
Cube = 18, Triple = (2,12,16)
Cube = 18, Triple = (9,12,15)
Cube = 19, Triple = (3,10,18)
Cube = 20, Triple = (7,14,17)
Cube = 24, Triple = (12,16,20)

AC代码：

#include<stdio.h>int main(){int n;while(scanf("%d",&n)!=EOF){if(n<6) break;for(int i=6;i<=n;i++)for(int j=2;j<=n;j++)for(int b=j;b<=n;b++)for(int d=b;d<=n;d++)if(i*i*i==j*j*j+b*b*b+d*d*d)printf("Cube = %d, Triple = (%d,%d,%d)\n",i,j,b,d);}return 0;}

内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理

标签：

相关文章推荐

新的分享

章节导航