Codeforces Round #190 (Div. 2)---A. Ciel and Dancing

Ciel and Dancing

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Fox Ciel and her friends are in a dancing room. There are n boys and m girls
here, and they never danced before. There will be some songs, during each song, there must be exactly one boy and one girl are dancing. Besides, there is a special rule:

either the boy in the dancing pair must dance for the first time (so, he didn't dance with anyone before);

or the girl in the dancing pair must dance for the first time.

Help Fox Ciel to make a schedule that they can dance as many songs as possible.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 100)
— the number of boys and girls in the dancing room.

Output

In the first line print k — the number of songs during which they can dance. Then in the following k lines,
print the indexes of boys and girls dancing during songs chronologically. You can assume that the boys are indexed from 1 to n, and the girls are indexed
from 1 to m.

Sample test(s)

input

2 1

output

2
1 1
2 1

input

2 2

output

3
1 1
1 2
2 2

Note

In test case 1, there are 2 boys and 1 girl. We can have 2 dances: the 1st boy and 1st girl (during the first song), the 2nd boy and 1st girl (during the second song).

And in test case 2, we have 2 boys with 2 girls, the answer is 3.

解题思路：n个boy，m个girl，若每对舞伴中至少有一个之前一次也没都跳过，问能够组成多少对舞伴，并输出。

贪心，再加上点数学。稍微动点数学常识就可以得出，最多可以组成 n+m-1 对满足要求的舞伴，然后就是怎么构造这么多对舞伴了。可以这样想，我们先用1号boy跟所有的

girl配对，然后再用剩下的n-1个boy分别跟最后一个girl配对即可。

AC代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff

int main()
{
    #ifdef sxk
        freopen("in.txt","r",stdin);
    #endif
    int n,m;
    while(scanf("%d%d",&n, &m)!=EOF)
    {
        printf("%d\n", m + n - 1);
        for(int i=1; i<=m; i++)
            printf("%d %d\n", 1, i);
        for(int j=2; j<=n; j++)
            printf("%d %d\n", j, m);
    }
    return 0;
}