A simple problem（湘大邀请赛）

A simple problem
Accepted : 61		Submit : 418
Time Limit : 15000 MS		Memory Limit : 655360 KB

Problem Description

There is a simple problem. Given a number N. you are going to calculate N%1+N%2+N%3+...+N%N.

Input

First line contains an integer T, there are T(1≤T≤50) cases. For each case T. The length N(1≤N≤10¹²).

Output

Output case number first, then the answer.

Sample Input

1
5

Sample Output

Case 1: 4

ps：http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1203

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <climits>
 #include <string>
 #include <map>
 #include <vector>
 #include <set>
 #include <list>
 #include <cstdlib>
 #include <cmath>
 #include <algorithm>
 #include <queue>
 #include <stack>
 #include <functional>
 #include <complex>
 #define mp make_pair
 #define X first
 #define Y second
 #define MEMSET(a, b) memset(a, b, sizeof(a))
 using namespace std;

 typedef unsigned int ui;
 typedef long long ll;
 typedef unsigned long long ull;
 typedef pair<int, int> pii;
 typedef vector<int> vi;
 typedef vi::iterator vi_it;
 typedef map<int, int> mii;
 typedef priority_queue<int> pqi;
 typedef priority_queue<int, vector<int>, greater<int> > rpqi;
 typedef priority_queue<pii> pqp;
 typedef priority_queue<pii, vector<pii>, greater<pii> > rpqp;

 const int MAX_N =  + ;
 const ll LL = ;
 int a[MAX_N];

 struct bigNum
 {
     ll bit[];

     bigNum() {
     }

     bigNum(const ll &b = ) {
         bit[] = b % LL;
         bit[] = b / LL;
         bit[] = ;
     }

     void format() {
         bit[] += bit[] / LL;
         bit[] %= LL;
         bit[] += bit[] / LL;
         bit[] %= LL;
     }

     bigNum operator * (const bigNum &bg) const {
         bigNum tmp();
         for (int i = ; i < ; ++i) {
             for (int j = ; j <= i; ++j) tmp.bit[i] += bit[j] * bg.bit[i - j];
         }
         tmp.format();
         return tmp;
     }

     void operator = (const bigNum &bg) {
         for (int i = ; i < ; ++i) bit[i] = bg.bit[i];
     }

     void operator *= (const bigNum &bg) {
         *this = *this * bg;
     }

     bigNum operator + (const bigNum &bg) const {
         bigNum tmp();
         for (int i = ; i < ; ++i) tmp.bit[i] = bit[i] + bg.bit[i];
         tmp.format();
         return tmp;
     }

     void operator += (const bigNum &bg) {
         *this = *this + bg;
     }

     void half() {
         if (bit[] % ) bit[] += LL;
         bit[] /= ;
         if (bit[] % ) bit[] += LL;
         bit[] /= ;
         bit[] /= ;
     }

     void print() {
         bool flag = false;
         if (bit[]) printf("%I64d", bit[]), flag = true;
         if (flag) printf("%09I64d", bit[]);
         else if (bit[]) printf("%I64d", bit[]), flag = true;
         if (flag) printf("%09I64d", bit[]);
         else printf("%I64d", bit[]);
     }
 };

 int main(int argc, char *argv[])
 {
 //    freopen("D:\\in.txt", "r", stdin);
     int t;
     cin >> t;
     for (int cas = ; cas <= t; ++cas) {
         ll n;
         scanf("%I64d", &n);
         bigNum bn(n), ans();
         int k = (int)sqrt((double)n);
         for (int i = ; i <= k; ++i) {
             bigNum tmp1(n / i - n / (i + ));
             bigNum tmp2(n + n - i * (n / i + n / (i + ) + ));
             tmp1 *= tmp2;
             tmp1.half();
             ans += tmp1;
         }
         ll tmp();
         int lmt = (int)(n / (k + ));
         for (int i = ; i <= lmt; ++i) tmp += n % i;
         bigNum tt(tmp);
         ans += tt;
         printf("Case %d: ", cas);
         ans.print();
         puts("");
     }
     return ;
 }