hdu 1061 Rightmost Digit

2019-04-14 16:44发布

//模幂运算,而且求个位,一直求10的模 # include # include # include using namespace std; __int64 Pow(__int64 m,__int64 n) { __int64 p=1; while(n) { if(n%2) p=p*m%10; n/=2; m=m*m%10; } return p%10; } int main() { int t; __int64 n,m; while(~scanf("%d",&t)) { while(t--) { scanf("%I64d",&n); m=n; m%=10;//求个位的n次取模 if(m==0) printf("0 "); else printf("%I64d ",Pow(m,n)); } } return 0; }