Codeforces 448D Multiplication Table（二分）

Multiplication TableTime Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d& %I64uSubmit StatusDescriptionBizon the Champion isn't just charming, he also is very smart.While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted ann × m multiplication table, where the element onthe intersection of the i-th row and j-th column equals i·j (therows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately.Can you repeat his success?Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-thnumber you write out is called the k-th largest number.InputThe single line contains integers n, m and k(1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m).OutputPrint the k-th largest number in a n × m multiplication table.Sample InputInput

2 2 2

Output

2

Input

2 3 4

Output

3

Input

1 10 5

Output

5

HintA 2 × 3 multiplication table looks like this:

1 2 32 4 6

1	2	3	4	5	6	7	8	9	10
2	4	6	8	10	12	14	16	18	20
3	6	9	12
4	8	12	16
5	10	15	20
6
7
8
9
10

如图给出表格前n行，前m，列，求第k个数

在1到n*m+1之间二分查找求出从小到大第k个数(这个题要注意不是第k大的数）对于每个数x,乘法表第i行都有min((m-1)/i,m)个数比x小,累加后与k比较即可#include<cstdio>#include<iostream>#include<algorithm>using namespace std;long long m,n,k;long long sum(long long m,long long n,long long k){long long sum=0;for(int i=1;i<=n;i++){sum+=min(k,(m-1)/i);}return sum;}int main(){while(~scanf("%lld%lld%lld",&n,&m,&k)){long long left=1,right=m*n+1;while(left<right){long long mid=(left+right)/2;if(sum(mid,n,m)<k)left=mid+1;elseright=mid;}printf("%lld\n",left-1);}return 0;}