題目描述

To facilitate the analysis of  a DNA sequence,  a DNA sequence is represented by a binary  number. The group of  DNA-1 has discovered a great new way .  There is a certain correlation between binary number and prime number. Instead of using the ordinary decadic numbers, they use prime base numbers. Numbers in this base are expressed as sequences of zeros and ones similarly to the binary numbers, but the weights of bits  in the representation are not powers of two, but the elements of the primes  ( 2, 3, 5, 7,... ).

For example  01101 , ie. 2+5+7=14

Herefore, it is necessary to convert the binary number to the sum of prime numbers 

輸入

The input consists of several instances, each of them consisting of a single line. Each line of the input contains a 01 string, length of not more than 150.  The end of input is not marked in any special way.

輸出

For each test case generate a single line containing a single integer , The sum of the primes.

樣例輸入

000010
0011
11001

樣例輸出

3
5
20

分析

程式碼

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
int prime[160]={0};
int isprime(int n)
{
    if(n==2)return 1;
    for(int i=2;i*i<=n;i++)
    {
        if(n%i==0) return 0;
    }
    return 1;
}
int main()
{
    int j=0;
    char s[160];
    for(int i=2;i<=1000;i++)
    {
        if(isprime(i)==1)
        {
            prime[j]=i;
            j++;
        }
    }//將從2開始的前160個素數存入陣列
    while(scanf("%s",s)!=EOF)
    {
        int sum=0,j=0;
        for(int i=strlen(s)-1;i>=0;i--,j++)//字串的倒序對應著素數陣列的順序
        {
            if(s[i]=='1')sum=sum+prime[j];//判斷此位置素數對應的權值是否為1，為1則加入總和
        }
        printf("%d\n",sum);
    }
}