#include <bits/stdc++.h> using namespace std; long long sum,ans=1,a,b; int main(){ cin>>a>>b; if(a==1){ cout<<1; return 0; } for(int i=0;i<b;i++){ ans*=a; if(ans>(long long)1e9){ cout<<"-1"; return 0; } } cout<<ans; return 0; }

能A

#include<bits/stdc++.h> using namespace std; int main(){ long long a,b,c,d=1000000000; cin>>a>>b; c=pow(a,b); if(c>d||c<0) cout<<-1; else cout<<c; return 0; }

呵呵，题目没说a和b会小于0

#include <bits/stdc++.h> using namespace std; int main(){ int a,b,sum=1; cin>>a>>b; if(pow(a,b)>1000000000){//判断是否超过10的9次方 cout<<"-1"; return 0; }else{ for(int i=0;i<b;i++){ sum*=a;//计算a的b次方 } } cout<<sum; return 0; }