90分求助

THU_ACMER @ 2024-09-18 08:32:57

import math
lst = input().split()
n = int(lst[0])
m = float(lst[1])
s = (n**2+n)//2
a = math.ceil(s*m*0.01)
if a > n or s*m*0.01 < 1:
    a = -1
print(a)

by shenyibo12200 @ 2024-10-12 20:44:08

@THU_ACMER

鄙人不才有点看不懂您写的Python代码，只能提供一下鄙人写的c++ ACcode

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int MANX=1e5;
double n;
double m;
int u=-1;
int main(){
    scanf("%lf%lf",&n,&m);
    double L=n/((1+n)*n/2);
    if(L<(m/100)){
        printf("%d",u);
        return 0;
    } 
    double ans=1;
    double l,q;
    for(double i=1;i<=n;i++){
        l=i/((1+n)*n/2);
        q=(i-1)/((1+n)*n/2);
        if(l>=(m/100)&&q<(m/100)){
            ans=i;
        } 
        else{
            continue;
        } 
    }
    printf("%.0lf",ans);
    return 0;
}