PAT 1029 Median (25 分)

#include
#include
using namespace std;

int main(){
    int n,m,cnt=0,ans,i,j;
    cin>>n;
    vector a1(n+1);
    for(i=1; i<=n;i++) scanf("%d", &a1[i]);
    cin>>m;
    vector a2(m+1);
    for(i=1; i<=m;i++) scanf("%d", &a2[i]);
    int target=(m+n+1)/2;
    i=1,j=1;
    while (i<=n&&j<=m){
        ans=a1[i] <= a2[j] ? a1[i++] : a2[j++];
        if(++cnt==target)break;
    }
    if(i <= n && cnt < target) 
    ans = a1[i + target - cnt - 1];
    else if(j <= m && cnt < target)
    ans = a2[j + target - cnt - 1];
    printf("%d", ans);
    system("pause");
    return 0;
}