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题意：计算从1开始到第n个非完全平方数的开方和
分析：设第n个非完全平方数的值为a，x * x < a < (x+1) * (x+1)，而且易得(tmp = sqrt (a) ) == x，a之前的非完全平方数的个数为a - tmp，所以可以二分查找a - tmp == n的a，然后模拟一下能计算出前a个数的开方和
收获：二分查找是个好方法
代码：
/************************************************
* Author :Running_Time
* Created Time :2015-8-27 16:14:57
* File Name :C.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int main(void) {
int T; scanf ("%d", &T);
while (T--) {
int n; scanf ("%d", &n);
ll l = 1, r = 1e12, ans1 = 0, ans2 = 0;
while (l <= r) {
ll mid = (l + r) >> 1;
ll tmp = sqrt (mid);
if (mid - tmp == n) {
ans1 = mid;
if (tmp * tmp == mid) ans1--;
break;
}
else if (mid - tmp > n) r = mid - 1;
else l = mid + 1;
}
ll tmp = sqrt (ans1);
for (ll i=1; i<tmp; ++i) {
l = i * i, r = (i + 1) * (i + 1);
ans2 += (r - l) * i;
}
ans2 += (ans1 - tmp * tmp + 1) * tmp;
printf ("%I64d %I64d\n", ans1, ans2);
}
return 0;
}
二分查找+数学 HDOJ 4342 History repeat itself
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原文地址：http://www.cnblogs.com/Running-Time/p/4765521.html
