UVALive 6862 Triples （找规律 暴力）

Triples

题目链接：

http://acm.hust.edu.cn/vjudge/contest/130303#problem/H

Description

http://7xjob4.com1.z0.glb.clouddn.com/4c05ed01e62e808388fc90e37554d588

Input

The input file contains several test cases, each of them as described below.
The first line contains the value of m and the second line contains the value of n.

Output

For each test case, the result will be written to standard output, on a line by itself.

Sample Input

```
85
95
```

Sample Output

```
8128
```

Source

2016-HUST-线下组队赛-4


##题意：

求有多少个三元组(x,y,z)满足 0

##题解：

由于n和m的数据都非常小，稍微打一下表就可以发现 j > 2 的时候不存在满足条件的三元组.
upd：实际上，上述表述为[费马大定理](http://baike.baidu.com/link?url=Ap0jvXEvOF9go2DTQ7LieopZRcrkcECDceCIu4VequB0IE9Bo3d8IC6gNq0yYnHmLDpfB9BQrWJRyS4xx8QYS9Yx7Lv0qJ_LvQifKep4zNe)
x=0时：y = z ∈ [0, m]. 故有 (m+1) * (n-1) 个解.
j=2时：暴力求一下即可.


##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define maxn 101000
#define inf 0x3f3f3f3f
#define mod 1000000007
#define mid(a,b) ((a+b)>>1)
#define eps 1e-8
#define IN freopen("in.txt","r",stdin);
using namespace std;

int solve(int m) {

int cnt = 0;

for(LL i=1; i<=m; i++) {

for(LL j=i; j<=m; j++) {

LL x = ii + jj;

for(LL k=0; k<=m; k++) {

if(k*k == x) {

//printf("%d %d %d\n",i,j,k);

cnt++;

break;

}

}

}

}

return cnt;

}

int main()

{

//IN;

int m,n;
while(scanf("%d %d", &m,&n) != EOF)
{
    int ans = solve(m) + (m+1)*(n-1);

    printf("%d\n", ans);
}

return 0;

}