UVA 11526 H(n) (数论)

题意：输入n（在32位带符号整数范围内），计算下面C++函数的返回值。

分析：合并同类项加速一下即可。

假设n为100，则n/34=2，那么通过n/2=50,可知道34到50的结果都是2，以此加速。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b) {
    if(fabs(a - b) < eps)  return 0;
    return a < b ? -1 : 1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 16384 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        LL n;
        scanf("%lld", &n);
        LL ans = 0;
        for(LL i = 1; i <= n; ++i){
            LL tmp = n / i;
            LL et = n / tmp;
            ans += (et - i + 1) * tmp;
            i = et;
        }
        printf("%lld\n", ans);
    }
    return 0;
}