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

 

 

       寻找&星空の孩子

 

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 = 100000 + 2;
const ll LL = 1000000000;
int a[MAX_N];

struct bigNum
{
    ll bit[3];
    
    bigNum() {
    }
    
    bigNum(const ll &b = 0) {
        bit[0] = b % LL;
        bit[1] = b / LL;
        bit[2] = 0;
    }
    
    void format() {
        bit[1] += bit[0] / LL;
        bit[0] %= LL;
        bit[2] += bit[1] / LL;
        bit[1] %= LL;
    }
    
    bigNum operator * (const bigNum &bg) const {
        bigNum tmp(0);
        for (int i = 0; i < 3; ++i) {
            for (int j = 0; j <= i; ++j) tmp.bit[i] += bit[j] * bg.bit[i - j];
        }
        tmp.format();
        return tmp;
    }
    
    void operator = (const bigNum &bg) {
        for (int i = 0; i < 3; ++i) bit[i] = bg.bit[i];
    }
    
    void operator *= (const bigNum &bg) {
        *this = *this * bg;
    }
    
    bigNum operator + (const bigNum &bg) const {
        bigNum tmp(0);
        for (int i = 0; i < 3; ++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[2] % 2) bit[1] += LL;
        bit[2] /= 2;
        if (bit[1] % 2) bit[0] += LL;
        bit[1] /= 2;
        bit[0] /= 2;
    }
    
    void print() {
        bool flag = false;
        if (bit[2]) printf("%I64d", bit[2]), flag = true;
        if (flag) printf("%09I64d", bit[1]);
        else if (bit[1]) printf("%I64d", bit[1]), flag = true;
        if (flag) printf("%09I64d", bit[0]);
        else printf("%I64d", bit[0]);
    }
};

int main(int argc, char *argv[])
{
//    freopen("D:\\in.txt", "r", stdin);
    int t;
    cin >> t;
    for (int cas = 1; cas <= t; ++cas) {
        ll n;
        scanf("%I64d", &n);
        bigNum bn(n), ans(0);
        int k = (int)sqrt((double)n);
        for (int i = 1; i <= k; ++i) {
            bigNum tmp1(n / i - n / (i + 1));
            bigNum tmp2(n + n - i * (n / i + n / (i + 1) + 1));
            tmp1 *= tmp2;
            tmp1.half();
            ans += tmp1;
        }
        ll tmp(0);
        int lmt = (int)(n / (k + 1));
        for (int i = 1; i <= lmt; ++i) tmp += n % i;
        bigNum tt(tmp);
        ans += tt;
        printf("Case %d: ", cas);
        ans.print();
        puts("");
    }
    return 0;
}