Problem 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…

Perfect Cubes Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14901   Accepted: 7804 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…

1.链接地址: http://bailian.openjudge.cn/practice/2810/ http://bailian.openjudge.cn/practice/1543/ http://poj.org/problem?id=1543 2.题目: Perfect Cubes Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 13190   Accepted: 6995 Description For hundr…

Perfect Cubes Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12595   Accepted: 6707 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…

Perfect Cubes Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 15302   Accepted: 7936 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…

zju对时间要求比较高,这就要求我们不能简单地暴力求解(三个循环搞定),就要换个思路:因为在循环时,已知a,确定b,c,d,在外重两层循环中已经给定了b和c,我们就不用遍历d,我们可以利用d^3=a^3-b^3-c^3来判断这个d. 看代码: #include <stdio.h> #include <math.h> int main() { ;n<=;++n) { ],b[],c[],count=,i,flag=; ); ;i<n;i++) { int ai= i*i*…

水题:求n^3 =  a^3 + b^3 + c^3 ;暴力即可 #include<iostream> using namespace std; int main(){ int n ; cin>>n; ; i <= n ; i++){ ; j <= n ; j++){ for(int k = j ; k <= n ; k++){ for(int l = k ; l <= n ; l++){ if(i*i*i==j*j*j+k*k*k+l*l*l) print…

题意:给定n个立方体,让你重新涂尽量少的面,使得所有立方体都相同. 析:暴力求出每一种姿态,然后枚举每一种立方体的姿态,求出最少值. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostrea…

http://codeforces.com/gym/101257/problem/A 把它固定在(0,0, 0)到(2, 2, 2)上,每次都暴力dfs检查,不会超时的,因为规定在这个空间上,一不行,就会早早退出. 这样写起来比较好写. #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #include <…

这道题明显是一道暴力. 暴力枚举每一个 \(a, b, c, d\) 所以我就写了一个暴力.每个 \(a, b, c, d\) 都从 \(1\) 枚举到 \(100\) #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #define line cout << endl using namespace st…