C - たくさんの数式 / Many Formulas
Time limit : 2sec / Memory limit : 256MB
Score : 300 points
Problem Statement
You are given a string S consisting of digits between 1
and 9
, inclusive. You can insert the letter +
into some of the positions (possibly none) between two letters in this string. Here, +
must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
- 1≤|S|≤10
- All letters in S are digits between
1
and9
, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Sample Input 1
Copy
125
Sample Output 1
Copy
176
There are 4 formulas that can be obtained: 125
, 1+25
, 12+5
and 1+2+5
. When each formula is evaluated,
- 125
- 1+25=26
- 12+5=17
- 1+2+5=8
Thus, the sum is 125+26+17+8=176.
Sample Input 2
Copy
9999999999
Sample Output 2
Copy
12656242944
题意:把一个数用不同的切割方法 把每种切割后的数累加
题解:用二进制转化,用&符号判断切割的位置
#include<bits/stdc++.h> using namespace std; int main() { string s; cin>>s; int n=s.size(); int m=1<<(n-1); long long ans=0; for(int i=0;i<m;i++) { long long t=s[0]-'0'; for(int j=0;j<n;j++) { if(j==n-1||i&1<<j) //切的方法 { ans+=t; t=0; if(j==n-1) break; } t*=10; t+=s[j+1]-'0'; } } cout<<ans<<endl; }