提交时间:2021-06-19 16:05:24
运行 ID: 20630
#include<iostream> #include<cmath> short k,n,b3; int ans; int main(){ scanf("%hd,%hd",&k,&n); ans=pow(2,n)-pow(2,k-1); if(ans<1000000) printf("%d",ans); else{ b3=ans%1000; while(ans>=1000) ans/=10; printf("%.03hd,%hd",b3,ans); } return 0; }