HDU4342 History repeat itself 二分

先二分找到这个非平方数，然后对一个一个区间进行求和的预处理。要注意边界问题。

代码如下：

#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

typedef long long Int;

Int N, num;

long long int sum[1000005];

Int bsearch(Int l, Int r) {
    Int mid, left;
    while (l <= r) {
        mid = (l + r) >> 1;
        left = mid - (int)floor(sqrt(double(mid)));
        if (N > left) {
            l = mid + 1;
        }
        else {
            r = mid - 1;
        }
    }
    return l;
}

void pre()
{
    int lim = 1 << 16;
    for (int i = 1; i <= lim; ++i) {
        sum[i] = sum[i-1]+1LL * (2*i+1) * i;
    }
}

int main()
{
    Int T, pos, s;
    long long int ans;
    pre();
    scanf("%I64d", &T);
    while (T--) {
        ans = 0;
        scanf("%I64d", &N);
        pos = bsearch(1, 1LL<<32);
        s = (int)floor(sqrt(double (pos)))-1;
        ans += (s+1)*(pos - (s+1)*(s+1)+1);
        printf("%I64d %I64d\n", pos, ans + sum[s]);
    }
    return 0;
}