整个题目的数据都要为long long

#include<stdio.h>

typedef long long LL;

LL n, m, k;

LL check(LL mid)
{
    LL res = 0;
    for(int i = 1;i <= n;i ++)
    res += ( m * i) <= mid ? m : mid / i;//关键点：枚举行，然后累加每行小于mid的数的个数
    return res;
}

int main()
{
    scanf("%lld %lld %lld",&n,&m,&k);

    LL l = 1,r = n * m;
    while(l<r)
    {
        LL mid = ( l + r ) >> 1;

        if(check(mid) >= k) r = mid;//二分的边界点就是k
        else l = mid + 1;

    }

    printf("%lld",l);
    return 0;
}

二分就是求满足某一个性质的边界问题！！！