核心算法是秦九韶算法和二分法，题解如下： #include <bits/stdc++.h> using namespace std; double m,y,s; int t; int out(double k) { printf("%.1f",k100); exit(0); } void solve(double l,double r) { double k=(l+r)/2,u=r-l; double a=m; if(u<0.0001) out(k); for(int i=1;i<=t;i++) a=a(1+k)-y; if(a>0) solve(l,k); if(a<0) solve(k,r); if(a==0) out(k); } int main() { cin>>m>>y>>t; solve(0,5); }

#include<bits/stdc++.h> #define ll long long using namespace std; int w0,w,m; bool check(double x){ double left=w0; for(int i=1;i<=m;i++){ left=left-(w-leftx); } if(left<=0)return 1; else return 0; } int main(){ cin>>w0>>w>>m; double l=0,r=3,mid=(l+r)/2; while(abs(r-l)>0.00001){ mid=(l+r)/2; if(check(mid))l=mid; else r=mid; } printf("%.1lf%",mid100); return 0; }