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

 

 

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <climits>
  5 #include <string>
  6 #include <map>
  7 #include <vector>
  8 #include <set>
  9 #include <list>
 10 #include <cstdlib>
 11 #include <cmath>
 12 #include <algorithm>
 13 #include <queue>
 14 #include <stack>
 15 #include <functional>
 16 #include <complex>
 17 #define mp make_pair
 18 #define X first
 19 #define Y second
 20 #define MEMSET(a, b) memset(a, b, sizeof(a))
 21 using namespace std;
 22 
 23 typedef unsigned int ui;
 24 typedef long long ll;
 25 typedef unsigned long long ull;
 26 typedef pair<int, int> pii;
 27 typedef vector<int> vi;
 28 typedef vi::iterator vi_it;
 29 typedef map<int, int> mii;
 30 typedef priority_queue<int> pqi;
 31 typedef priority_queue<int, vector<int>, greater<int> > rpqi;
 32 typedef priority_queue<pii> pqp;
 33 typedef priority_queue<pii, vector<pii>, greater<pii> > rpqp;
 34 
 35 const int MAX_N = 100000 + 2;
 36 const ll LL = 1000000000;
 37 int a[MAX_N];
 38 
 39 struct bigNum
 40 {
 41     ll bit[3];
 42 
 43     bigNum() {
 44     }
 45 
 46     bigNum(const ll &b = 0) {
 47         bit[0] = b % LL;
 48         bit[1] = b / LL;
 49         bit[2] = 0;
 50     }
 51 
 52     void format() {
 53         bit[1] += bit[0] / LL;
 54         bit[0] %= LL;
 55         bit[2] += bit[1] / LL;
 56         bit[1] %= LL;
 57     }
 58 
 59     bigNum operator * (const bigNum &bg) const {
 60         bigNum tmp(0);
 61         for (int i = 0; i < 3; ++i) {
 62             for (int j = 0; j <= i; ++j) tmp.bit[i] += bit[j] * bg.bit[i - j];
 63         }
 64         tmp.format();
 65         return tmp;
 66     }
 67 
 68     void operator = (const bigNum &bg) {
 69         for (int i = 0; i < 3; ++i) bit[i] = bg.bit[i];
 70     }
 71 
 72     void operator *= (const bigNum &bg) {
 73         *this = *this * bg;
 74     }
 75 
 76     bigNum operator + (const bigNum &bg) const {
 77         bigNum tmp(0);
 78         for (int i = 0; i < 3; ++i) tmp.bit[i] = bit[i] + bg.bit[i];
 79         tmp.format();
 80         return tmp;
 81     }
 82 
 83     void operator += (const bigNum &bg) {
 84         *this = *this + bg;
 85     }
 86 
 87     void half() {
 88         if (bit[2] % 2) bit[1] += LL;
 89         bit[2] /= 2;
 90         if (bit[1] % 2) bit[0] += LL;
 91         bit[1] /= 2;
 92         bit[0] /= 2;
 93     }
 94 
 95     void print() {
 96         bool flag = false;
 97         if (bit[2]) printf("%I64d", bit[2]), flag = true;
 98         if (flag) printf("%09I64d", bit[1]);
 99         else if (bit[1]) printf("%I64d", bit[1]), flag = true;
100         if (flag) printf("%09I64d", bit[0]);
101         else printf("%I64d", bit[0]);
102     }
103 };
104 
105 int main(int argc, char *argv[])
106 {
107 //    freopen("D:\\in.txt", "r", stdin);
108     int t;
109     cin >> t;
110     for (int cas = 1; cas <= t; ++cas) {
111         ll n;
112         scanf("%I64d", &n);
113         bigNum bn(n), ans(0);
114         int k = (int)sqrt((double)n);
115         for (int i = 1; i <= k; ++i) {
116             bigNum tmp1(n / i - n / (i + 1));
117             bigNum tmp2(n + n - i * (n / i + n / (i + 1) + 1));
118             tmp1 *= tmp2;
119             tmp1.half();
120             ans += tmp1;
121         }
122         ll tmp(0);
123         int lmt = (int)(n / (k + 1));
124         for (int i = 1; i <= lmt; ++i) tmp += n % i;
125         bigNum tt(tmp);
126         ans += tt;
127         printf("Case %d: ", cas);
128         ans.print();
129         puts("");
130     }
131     return 0;
132 }

 

 

 

 

posted @ 2015-06-11 00:53  寻找&星空の孩子  阅读(304)  评论(0)  编辑  收藏  举报