3533. 庙会

include

int main()
{
int m, n, k, i;
scanf(“%d %d %d”, &m, &n, &k);
for (i=0; i<k; i++)
{
printf(“%d %d\n”, i%m+1, i%n+1);
}
}

根本不需要循环队列，搞两个队列，每次
int a = nan.front();nan.pop();nan.push(a);
int b = nv.front();nv.pop();nv.push(b);
cout<<a<<” “<<b<<endl;
就这么简单

看题目还以为是匈牙利

include

include

include

include

include

include

include

include

define isyeap(x) x%4==0 &&x%100!=0 || x%400==0 ? 1 : 0

using namespace std;
//priority_queue,greater\ > Q;// 优先队列，解决哈夫曼树问题

int main()
{
int m,n,k;
while(scanf(“%d%d%d”,&m,&n,&k)!=EOF)
{
int i=1,j=1;
while(k–)
{
printf(“%d %d\n”,i,j);
i++;
j++;
if(i>m) i=1;
if(j>n) j=1;
}
}
return 0;
}

include

int main(){
int m, n, k;
scanf(“%d%d%d”,&m,&n,&k);
int x=1, y=1, i;
for(i=0; im){
x=1;
}
printf(“%d\n”,y);
y++;
if(y>n){
y=1;
}
}
}

提示：该问题的解决应该是采用类似“循环队列”的方式。