Uva 11881 - Internal Rate of Return

你为何这么叼！

Internal Rate of Return

Time limit: 1.000 seconds

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 + $\displaystyle {CF_1 \over {1+IRR}}$ + $\displaystyle {CF_2 \over {(1+IRR)^2}}$ + K + $\displaystyle {CF_T \over {(1+IRR)^T}}$ = 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 ( 1 $\le$ T $\le$ 10), 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 no IRR exists, print ``No" (without quotes); if there are multiple IRRs, print ``Too many"(without quotes).

Sample Input

Sample Output

1.00
0.50

Problemsetter: Rujia Liu, Special Thanks: Yiming Li, Shamim Hafiz & Sohel Hafiz

 1 #pragma comment(linker, "/STACK:1024000000,1024000000")
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <iostream>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <queue>
 8 #include <set>
 9 #include <map>
10 #include <string>
11 #include <math.h>
12 #include <stdlib.h>
13 #include <time.h>
14 using namespace std;
15 #define maxn 101
16 #define INF 0x7fffffff
17 int n;
18 int a[maxn];
19 double fun(double m){
20     double s=0.0,x=1.0;
21     for(int i=1;i<=n;i++){
22         x*=(m+1);
23         s+=a[i]/x;
24     }
25     return s;
26 }
27 double find(){
28     double l=-1.0,r=INF*1.0,m;
29     for(int i=0;i<=100;i++){
30         m=l+(r-l)/2;
31         if(fun(m)>=-a[0])l=m;
32         else r=m;
33     }
34     return m;
35 }
36 int main(){
37     while(scanf("%d",&n)&&n){
38         int k=0;
39         for(int i=0;i<=n;i++){scanf("%d",&a[i]);if(a[i]>0)k++;}
40         if(k==n+1){printf("No\n");continue;}
41         printf("%.2lf\n",find());
42     }
43     return 0;
44 }

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posted @ 2013-10-09 15:00 HaibaraAi 阅读(249) 评论(0) 收藏举报

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