UPC——神仙贷款—>二分

#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
///#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
///char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
const double limit=1e-6;///在后面有用到二分的精度问题在这里设为常量
#define start int wuyt()
#define end return 0
ll num[20008];
ll num2[20008];
int n,m,cnt;
ll gcd(ll a,ll b)
{
    ll t;
    while(b!=0)
    {
        t=a%b;
        a=b;
        b=t;
    }
    return a;
}
ll ans;
ll qPow(ll x, ll k)
{
    ll res = 1;
    while(k) {
        if(k&1)
      res=(res*x);
        k>>=1;
        x=(x*x);
    }
    return res;
}
bool check(double number){
    double temp=0;
    for(int i=1;i<=cnt;i++)
        temp+=n/pow(1+number,i);
    if(temp>=m) return 1;
    else return 0;
}
///ll a[40][40];
ll a,b;
char ss[50];
int pos;
bool judge(ll num[]){
    for(int i=1;i<=2*n;i++){
        if(i!=num[i])
            return false;
    }
    return true;
}
void work(ll num[],ll num2[]){
    int cnt=1;
    for(int i=2*n;i>n;i--){
        num2[i-cnt]=num[i];
        cnt++;
    }
    for(int i=1;i<=n;i++){
        num2[2*i]=num[i];
    }
    for(int i=1;i<=2*n;i++) num[i]=num2[i];
}
int main()
{
    m=read,n=read,cnt=read;
    double left=0,right=100;
    double mid;
    while(right-left>limit){
        mid=(left+right)/2;
        if(check(mid)) left=mid;
        else right=mid;
    }
    printf("%.1f\n",left*100);
    return 0;
}

bool check(double number){
    double temp=0;
    for(int i=1;i<=cnt;i++)
        temp+=n/pow(1+number,i);
    if(temp>=m) return 1;
    else return 0;
}