Problem C. Jane's Flower Shop Google APAC 2017 University Test Round A

这一题一开始想的是求导找单调性，最后没发现什么就呵呵了。把IRR方程左边表达式记为f(r)。之后得知，因为方程f(r)=0在[0,1]有且只有一个解，并且在-1点处是正值，因此可以通过二分查找得出方程的根。

这一题对精度要求很高。之前二分查找中判断f(r)=0的精度设低就WA了==。如果二分查不到f(r)=0的点，也要返回当前mid值。虽然这个值确实是根，可f(mid)就是和0差得很大。并不造原因。Plus，我第一次得知还有long double 类型。o(╯□╰)o

#include<iostream>#include<stdio.h>#include<cstdio>#include<string>#include<cmath>#include<stdlib.h>#include<algorithm>#include<string.h>#include<cstring>#include<vector>#include<queue>#include<map>using namespace std;//2017 RoundA Problem C. Jane's Flower Shopint T;const int maxn=110;const double eps=1e-10;//use 1e-9 leads to WAint M;int C[maxn];double ans;long double fun(double r){    long double power[maxn];    memset(power,0,sizeof(power));    power[0]=1;    for(int i=1;i<=M;i++)    {        power[i]=power[i-1]*(1+r);        //cout<<power[i]<<endl;    }    long double ret=0;    for(int i=0;i<M;i++)    {        //cout<<i<<" "<<C[M-i]<<" "<<power[i]<<" "<<C[M-i]*power[i]<<endl;        ret+=C[M-i]*power[i];//pow(1+r,i);    }    //cout<<ret<<endl;    //cout<<0<<" "<<C[0]<<" "<<power[M]<<" "<<C[0]*power[M]<<endl;    ret-=C[0]*power[M];//pow(1+r,M);    return ret;}double binartsearch(){    double left=-1;    double right=1;    double mid=(left+right)/2;    while(left+eps<right)    {        mid=(left+right)/2;        long double tmp=fun(mid);        //cout<<left<<" "<<right<<" "<<mid<<" "<<tmp<<" "<<fun(0.618034)<<endl;        if(abs(tmp)<1e-13)//set <eps leads to WA        {            return mid;        }        if(tmp>0)        {            left=mid;//+eps;        }        else if(tmp<0)        {            right=mid;//-eps;        }    }    return mid;//there must be a solution, then return mid is fine.}int main(){    freopen("C-small-practice.in","r",stdin);//    freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout);    scanf("%d",&T);    for(int ca=1;ca<=T;ca++)    {        scanf("%d",&M);        memset(C,0,sizeof(C));        ans=0;        for(int i=0;i<=M;i++)        {            scanf("%d",&C[i]);        }        ans=binartsearch();        printf("Case #%d: %.12f\n",ca,ans);    }    return 0;}