lightoj 1008

水题，开根号判断大致范围，再找即可。

#include<cstdio>
#include<cmath>
#include<cstdlib>
using namespace std;
int main(){
    int t, CASE(0);
    long long int n;
    scanf("%d", &t);
    while(t--){
        scanf("%lld", &n);
        long long int m = sqrt(n);
        printf("Case %d: ",++CASE);
        if(m*m == n){
            if(m&1) printf("1 %lld\n", m);
            else printf("%lld 1\n", m);
            continue;
        }
        if(m&1){
            if(m*m+m+1 >= n) printf("%lld %lld\n", n-m*m, m+1);
            else printf("%lld %lld\n", m+1, (m+1)*(m+1)-n+1);
        }else{
            if(m*m+m+1 >= n) printf("%lld %lld\n", m+1, n-m*m);
            else printf("%lld %lld\n", (m+1)*(m+1)-n+1, m+1);
        }
    }
}