A - A

A - A
Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u
Submit
 
Status
Description
For 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 program to find all sets of numbers {a,b,c,d} which satisfy this equation for a <= N.
Input
One integer N (N <= 100).
Output
The 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 exist several 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
24
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)

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代码：

#include<cstdio>#include<cstring>struct node{int sh;int a[10],b[10],c[10];}shu[102];int p[102];bool fafe(int ii){bool ka=false;for (int i=2;i<ii-2;i++)for (int j=i+1;j<ii-1;j++)for (int s=j+1;s<ii;s++)if ((p[i]+p[j]+p[s])==p[ii]){ka=true;shu[ii].a[shu[ii].sh]=i;shu[ii].b[shu[ii].sh]=j;shu[ii].c[shu[ii].sh]=s;shu[ii].sh++;}return ka;}int main(){int t;scanf("%d",&t);for (int i=1;i<=t;i++)p[i]=i*i*i;memset(shu,0,sizeof(shu));for (int i=1;i<=t;i++){if (fafe(i)){for (int j=0;j<shu[i].sh;j++){   // printf("1213341\n");printf("Cube = %d, Triple = (%d,%d,%d)\n",i,shu[i].a[j],shu[i].b[j],shu[i].c[j]);}}}return 0; }