UVA 11881 Internal Rate of Return（数学+二分）

In finance, Internal Rate of Return (IRR) is the discount rate of an investment when NPV equals zero. Formally, given T, CF₀, CF₁, ..., CF_T, then IRR is the solution to the following equation:

NPV = CF₀ +

+ K +

= 0

Your task is to find all valid IRRs. In this problem, the initial cash-flow CF₀ < 0, while other cash-flows are all positive (CF_i > 0 for all i = 1, 2,...).

Important: IRR can be negative, but it must be satisfied that IRR > - 1.

Input

There will be at most 25 test cases. Each test case contains two lines. The first line contains a single integer T ( 1T10), the number of positive cash-flows. The second line contains T + 1 integers: CF₀, CF₁,CF₂, ..., CF_T, where CF₀ < 0, 0 < CF_i < 10000 ( i = 1, 2,..., T). The input terminates by T = 0.

Output

For each test case, print a single line, containing the value of IRR, rounded to two decimal points. If noIRR exists, print ``No" (without quotes); if there are multiple IRRs, print ``Too many"(without quotes).

题目大意：给出CF[0]<0，CF[i]>0，i>0，求IRR（IRR>-1）令NPV = 0.

思路：设f(IRR) = NPV，这就变成了一个函数，稍微观察一下，可以发现当IRR∈(-1, +∞)的时候是单调递减的（好像是吧做完忘了），这样我们就可以二分答案0点了。当IRR无限接近-1的时候，f(IRR)→+∞（好像是吧），当IRR→+∞时，f(IRR)→CF[0]<0，令left = -1、right = 1e5（我也不知道该取什么我随便取的然后AC了），随便二分一下就好。

PS：恩？说完了？那什么时候输出No和Too many啊？关于这个坑爹的问题，看完前面的分析，笔者完全不知道什么时候会出现这两个答案，于是妥妥地没将这两个东西写进代码然后AC了。这里还有一个小技巧，题目的样例完全没有出现No和Too many这两种答案，很可能说明这两种答案都是不存在的。比如很多的题目说如果什么什么得不到答案就输出-1那样，它的样例大多都会有一个是输出-1的，当然这不是绝对的……

代码（15MS）：

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 #include <iostream>

 #include <cmath>

 using namespace std;

 const int MAXN = ;

 const double EPS = 1e-;

 inline int sgn(double x) {

     if(fabs(x) < EPS) return ;

     return x >  ?  : -;

 }

 int CF[MAXN];

 int T;

 double f(double IRR) {

     double ret = CF[], tmp =  + IRR;

     for(int i = ; i <= T; ++i) {

         ret += CF[i] / tmp;

         tmp = tmp * ( + IRR);

     }

     return ret;

 }

 double solve() {

     double ans = -;

     double L = -, R = 1e5;

     while(R - L > EPS) {

         double mid = (R + L) / ;

         if(sgn(f(mid)) == ) L = mid;

         else R = mid;

     }

     return ans = L;

 }

 int main() {

     while(scanf("%d", &T) != EOF) {

         if(T == ) break;

         for(int i = ; i <= T; ++i) scanf("%d", &CF[i]);

         //double t; while(cin>>t) cout<<f(t)<<endl;

         printf("%.2f\n", solve());

     }

 }