可以用两个队列来模拟

#include <bits/stdc++.h>
using namespace std;

typedef queue<int> qi;
qi a,b;

int main()
{
	int m,w,s;
	cin >> m >> w >> s;
	for(int i=1;i<=m;i++) a.push(i);
	for(int j=1;j<=w;j++) b.push(j);
	while(s--)
	{
		cout << a.front() << " " << b.front() << endl;
		a.push(a.front());
		b.push(b.front());
		a.pop();
		b.pop();
	}
	return 0;
}

#include <iostream>

using namespace std;

int main()
{
    int m,w,i;
    cin>>m>>w>>i;
    int j=1,s=1;
    for(int o=0;o<i;o++){
        cout<<j<<" "<<s<<endl;
        j++;
        s++;
        if(j>m) j=1;
        if(s>w) s=1;

    }
    return 0;
}

大道至简

l=[int(input()) for i in range(3)]
for i in range(l[2]):
    print(i%l[0]+1,i%l[1]+1)

n=int(input())
m=int(input())
num=int(input())
nan=[0]*n
nv=[0]*m
head1,tail1=0,0
head2,tail2=0,0
for i in range(1,n+1):
    nan[tail1]=i
    tail1+=1
for j in range(1,m+1):
    nv[tail2]=j
    tail2+=1
for k in range(num):
    print(nan[head1],nv[head2])
    head1=(head1+1)%n
    head2=(head2+1)%m

C++

思维简单，就是配对，哪边人数到了就从头来；

#include
using namespace std;
int main(){
    int m ,f; //male和female不用解释吧
    int n;
    int j = 1, k = 1;
    cin >> m >> f;
    cin >> n;
    for(int i = 1; i <= n; i++) {
        cout << j << " " << k; //配对输出
        cout << endl;
        j += 1; //这个male跳过舞了，+1!
        k += 1; //这个female也跳过了，+1 too!
        if(j == m + 1) j = 1; //如果所有male都跳过了，从头来
        if(k == f + 1) k = 1; //如果所有female都跳过了，也从头来
    }
    return 0;
}

Python

python就不用多说了吧

m = int(input())
f = int(input())
n = int(input())
j = 1
k = 1
for i in range(0, n):
    print("%d %d" %(j, k)) #配对输出
    j += 1 #这个male跳过舞了，+1!
    k += 1 #这个female也跳过了，+1 too!
    if j == m + 1: #如果所有male都跳过了，从头来
        j = 1
    if k == f + 1: #如果所有female都跳过了，也从头来
        k = 1

题解仅提供思路

n=int(input()) m=int(input()) num=int(input()) for i in range(num): print(i%n+1,i%m+1)

┭┮﹏┭┮

我的最麻烦

#include<bits/stdc++.h>

using namespace std;



int main()
{

    int a,b,c;
    cin >>a>>b>>c;
    int n[a+1];
    int m[b+1];
    for(int i=1; i<=a; i++){
        n[i]=i;
    }
    for(int i=1; i<=b; i++){
        m[i]=i;
    }
    int u,v;
    for(int i=1; i<=c; i++){
        cout<<n[1]<<" "<<m[1]<<endl;
        u=n[1];
        v=m[1];
        for(int j=1; j<a; j++){
            n[j]=n[j+1];
        }
        for(int j=1; j<b; j++){
            m[j]=m[j+1];
        }
        n[a]=u;
        m[b]=v;
    }

	return 0;
}

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int a,b,c,i=1;
    scanf("%d",&a);
    scanf("%d",&b);
    scanf("%d",&c);
    int man=1,woman=1;
    while (i<=c){
    printf ("%d %d\n",man,woman);
    
    man++; woman++;
    if (man>a) man=1;
    if (woman>b) woman=1;
    i++;
    }
    return 0;
}

m=int(input())

w=int(input())

song=int(input())

maxx=max(m,w,song)

man=[0]*(maxx+20)

woman=[0]*(maxx+20)

for i in range(1,m+1):

man[i-1]=i

headm=0

tailm=m

for j in range(1,w+1):

woman[j-1]=j

headw=0

tailw=w

k=0

while k<song:

print(man[headm],woman[headw])

man[tailm]=man[headm]

woman[tailw]=woman[headw]

headm+=1

tailm+=1

headw+=1

tailw+=1

k+=1