九度OJ 1095 2的冪次方
阿新 • • 發佈：2019-01-30
- 題目描述：
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Every positive number can be presented by the exponential form.For example, 137 = 2^7 + 2^3 + 2^0。
Let's present a^b by the form a(b).Then 137 is presented by 2(7)+2(3)+2(0). Since 7 = 2^2 + 2 + 2^0 and 3 = 2 + 2^0 , 137 is finally presented by 2(2(2)+2 +2(0))+2(2+2(0))+2(0).
Given a positive number n,your task is to present n with the exponential form which only contains the digits 0 and 2.
- 輸入：
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For each case, the input file contains a positive integer n (n<=20000).
- 輸出：
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For each case, you should output the exponential form of n an a single line.Note that,there should not be any additional white spaces in the line.
- 樣例輸入：
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1315
- 樣例輸出：
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2(2(2+2(0))+2)+2(2(2+2(0)))+2(2(2)+2(0))+2+2(0)
本題從輸出的結果就可以看出來，這是一道遞迴的例子。對於當前的數，先確定它在2^i次方左右，這裡的i先確定下來,即使得2^i<=n成立的最大的i,然後繼續求n-pow(2,i),直到n=0 對於指數，也是同樣的程式碼，只不過這裡要注意的是，遞迴的出口，當指數為0，為1，為2時，作為最基本的情況，要單獨列出。#include <iostream> #include <math.h> using namespace std; void present(int n){ if(n == 0) cout<<"0"; if(n == 1){ return; } if(n == 2){ cout<<"2"; return; } int i = 0; int j = 0; while(n){ for(i=0;i<18;i++){ if(pow(2,i) > n) break; } j = i - 1; if(j!=1) cout<<"2("; else cout<<"2"; present(j); if(j!=1) cout<<")"; n -= pow(2,j); if(n != 0) cout<<"+"; } } int main(){ int n; while(cin>>n){ present(n); cout<<endl; } return 0; }