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POJ 3685:Matrix 二分

2015-10-29 11:41 120 查看
Matrix

Time Limit: 6000MSMemory Limit: 65536K
Total Submissions: 5489Accepted: 1511
Description

Given a N × N matrix A, whose element in the i-th row and j-th column Aij is an number that equals i2 + 100000 × i + j2 - 100000 × j + i × j,
you are to find the M-th smallest element in the matrix.

Input

The first line of input is the number of test case.

For each test case there is only one line contains two integers, N(1 ≤ N ≤ 50,000) and M(1 ≤ M ≤ N × N). There is a blank line before each test case.

Output

For each test case output the answer on a single line.

Sample Input

12

1 1

2 1

2 2

2 3

2 4

3 1

3 2

3 8

3 9

5 1

5 25

5 10

Sample Output

3
-99993
3
12
100007
-199987
-99993
100019
200013
-399969
400031
-99939


给了一个N*N的矩阵,每个位置上的数由该位置的下标i,j决定。然后问这个矩阵中第m小的数。

通过这道题好好总结了一下二分,总算是深刻理解了一下。

第一个二分枚举答案,第二个二分,通过公式可知,函数跟j不是单调的关系,但跟i是单调的关系,所以每次枚举j,然后二分i的值。

代码:

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

#define maxn 1e12
typedef long long ll;

ll m, n;
ll le, ri, mid;

ll cal(ll i, ll j)
{
return i*i + 100000 * i + j*j - 100000 * j + i*j;
}

ll check(ll x)
{
ll i, le, ri, mid, cnt;

ll temp;
cnt = 0;
for (i = 1; i <= n; i++)
{
le = 1;//mid不能取到0,所以这里的le取1
ri = n + 1;//因为mid可能要取到n,所以这里的ri要取到比n大的数

mid = le + (ri - le) / 2;
while (le < ri)
{
temp = cal(mid, i);
if (cal(mid, i) < x)
{
le = mid + 1;
}
else
{
ri = mid ;
}
mid = le + (ri - le) / 2;
}
cnt += (mid - 1);//经过计算,这里始终是多计算了一个1,所以要在这里把它扣掉
}
return cnt;
}

int main()
{
//freopen("i.txt", "r", stdin);
//freopen("o.txt", "w", stdout);

int test;
scanf("%d", &test);

while (test--)
{
scanf("%lld%lld", &n, &m);
le = -maxn;
ri = maxn;

mid = le + (ri - le) / 2;
while (le < ri)
{
if (check(mid) < m)//检查小于 mid 的个数
{
le = mid + 1;
}
else
{
ri = mid;
}
mid = le + (ri - le) / 2;
}
printf("%lld\n", mid-1);//有m个小于mid的数,所以第m大的数就是mid-1
}

//system("pause");
return 0;
}
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