坑点!
由于m是double类型,double类型-1要保证精度问题,应该要减一个很小的数
#include#include using namespace std; typedef long long ll; int main() { cin.tie(0),cout.tie(0); int t; cin>>t; while(t--) { int n; double m; cin>>n>>m; ll res=0; ll red=0,green=0; while(n) { red+=n*100; green+=min((double)10000,n*100*(m-0.999999999)); res+=red/10+green/10; n=red/200; red=0,green=0; } cout<



