标签：Cubes int Sum number long two integer YES
一、题目
You are given a positive integer x. Check whether the number x is representable as the sum of the cubes of two positive integers.
Formally, you need to check if there are two integers a and b (1≤a,b) such that a^3 + b^3=x.
For example, if x=35, then the numbers a=2 and b=3 are suitable (2^3 + 3^3=8+27=35). If x=4, then no pair of numbers a and b is suitable.
两个数能否表示为立方和？
二、输入
The first line contains one integer t (1≤t≤100) — the number of test cases. Then t test cases follow.
Each test case contains one integer x (1≤x≤1e12).
Please note, that the input for some test cases won’t fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
第一行测试样例数t
每个样例一个数x
三、输出
YES/NO
四、样例
input:
7
1
2
4
34
35
16
703657519796
output:
NO
YES
NO
NO
YES
YES
YES
Note
The number 1 is not representable as the sum of two cubes.
The number 2 is represented as 1^3 + 1^3.
The number 4 is not representable as the sum of two cubes.
The number 34 is not representable as the sum of two cubes.
The number 35 is represented as 2^3 + 3^3.
The number 16 is represented as 2^3 + 2^3.
The number 703657519796 is represented as 5^7793 + 7^9933.
五、思路 + 代码
打表然后二分查找
不打表直接暴力，或者打表不用二分都会超时，不要问我是怎么知道的…
数据很大要用long long，打表只需打到1e4，打表时遍历用的i要用long long，这种低级错误恐怕只有我会犯了
代码如下：
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
ll g[10004], x;
int main() {
int t;
cin >> t;
for(ll i = 1; i <= 10000; i++) g[i] = i * i * i;
while (t--) {
cin >> x;
int f = 0;
for(int i = 1; g[i] < x / 2 + 1; i++) {
ll tmp = x - g[i];
int l = 1, r = 10000;
while (l < r) {
int mid = (l + r) >> 1;
if (g[mid] < tmp) {
l = mid + 1;
}
else {
r = mid;
}
}
if (g[l] == tmp) {
f = 1;
break;
}
}
if (f) cout << "YES" << endl;
else cout << "NO" << endl;
}
return 0;
}
标签：Cubes,int,Sum,number,long,two,integer,YES 来源： https://blog.csdn.net/qq_43697906/article/details/113833180
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