ural Infernal Work

Infernal Work

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Description

Railwaymen Vassily and Pyotr died and were sent to Hell. Their first punishment was to perform a complete inspection of the Moscow–Vladivostok railroad. They spent many weeks walking along the railroad together, one of them along the left rail and the other along the right rail, writing the long serial numbers of ties to their thick notebooks. As soon as they finished that infernal task, they immediately got a new task, which was even more meaningless. Now they had to count the number of pairs of ties that were written in Vassily's notebook on the same page and in Pyotr's notebook on different pages.

The friends came to you in a dream and asked you to save them from that terrible torment.

Input

The only input line contains integers a, b, n (1 ≤ a, b ≤ n ≤ 25 000 000). One page in Vassily's notebook comprises a numbers of ties, and one page in Pyotr's notebook comprises b numbers of ties. They have written numbers of n ties. All these numbers are different and are written in their notebooks in the same order.

Output

Output one number, which is the answer to the problem.

Sample Input

input	output
3 4 10	4
2 4 10	0

Notes

Let the ties in the first sample be numbered by the letters from A to J. Then the following four pairs satisfy the condition: (D, E), (D, F), (G, I), (H, I).

 

 

首先看第一种方法（超出空间限制）：

#include <iostream>
using namespace std;

struct Data{
    long long  zua;
    long long  zub;
};
//本题的思路是为两种不同的分组方式分别赋予不同的组别值，
//然后搜索符合题目要求的配对。 
int main(){
    long long  a,b,n;
    cin>>a>>b>>n;
    Data data[n];
      
    long long num1=1,num2=1;
    for(long long  i=0;i<n;i++){
        data[i].zua=num1;
        data[i].zub=num2;
        if((i+1)%a==0){
            ++num1;
        }
        if((i+1)%b==0){
            ++num2;
        }
    } 
    long long  result=0;
    for(long long  i=0;i<n;i++){
        for(long long  j=i;j<n&&j<i+a;j++){
            if(data[i].zua==data[j].zua&&data[i].zub!=data[j].zub){
                ++result;
            }
        }
    }
    cout<<result;
}
 

那么就用一个数组来实现它：

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

int main(){
    int a,b,n;
    cin>>a>>b>>n;
    long long result=0;
    bool qwe; 
    for(int i=0;i<n;i=i+a){
        int yb=i/b,k=1,cur;
        qwe=0;
        
        for(int j=i+1;j<i+a&&j<n;j++){
            int yb2=j/b;
            if(yb2==yb) continue;
            if(!qwe){
                cur=yb2;
                k=j-i;
                qwe=1;
            }
            if(yb2!=cur){
                k=j-i;
                cur=yb2;
            }
            result+=k;
        }
    }
    cout<<result<<endl;
}