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Crawling in process... Crawling failed Time Limit:6000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
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 题目大意：给你一个n*n矩阵，每一个位置都有一个值，这个值由该点的该点的行列标决定，问你第m小的元素是多少。
思路分析：比赛时都贴上二分标签了，最后还没敲出来，首次我们要二分答案，然后check，check函数里面按列
进行枚举，因为在j确定的情况下函数关于i递增，然后直接二分查找刚好小于等于mid的元素在每一列的位置，累加
return，判断返回的数与m的关系，继续二分
代码：
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <cmath> using namespace std; const int inf=1e9; typedef long long ll; const int C=100000; ll f(ll x,ll y) { return (x*x+C*x+y*y-C*y+x*y); } ll n,m; ll checknum(ll num) { ll cnt=0; for(int i=1;i<=n;i++)//枚举列，因为在列数确定的情况下表达式关于i是递增的 { ll sl=1,sr=n; int ant=0; while(sl<=sr) { ll smid=(sl+sr)>>1; if(f(smid,i)<=num) { ant=smid; sl=smid+1; } else sr=smid-1; } cnt+=ant; } return cnt; } int main() { int T; scanf("%d",&T); while(T--) { scanf("%lld%lld",&n,&m); ll l=-100000*n,r=n*n+100000*n+n*n+n*n; ll ans; while(l<=r) { ll mid=(l+r)>>1; //cout<<mid<<endl; if(checknum(mid)>=m) { ans=mid; //cout<<ans<<endl; r=mid-1; } else l=mid+1; // cout<<ans<<endl; } printf("%lld\n",ans); } }
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原文地址：http://www.cnblogs.com/xuejianye/p/5691120.html
