2.3.3 Zero Sum

dfs，模拟即可。不难

   
     

    
      
/*
    
      
ID: awsd1231
    
      
PROG: zerosum
    
      
LANG: C++
    
      
*/
    
      
#include<iostream>
    
      
#include<cstdio>
    
      
#include<string>
    
      

    
      
using namespace std;
    
      

    
      
int n;
    
      
char ch[] = {'+', '-', ' '};
    
      

    
      
void dfs(int x, int change, int befVal, string tmpS) {
    
      
 char tc = x + '0';
    
      
 switch (change) {
    
      
 case 0 : tmpS += "+"; tmpS += tc; befVal += x; break;
    
      
 case 1 : tmpS += "-"; tmpS += tc; befVal -= x; break;
    
      
 case 2 : char tmp = '+'; int tmpVal = 0;
    
      
 for (int i = 0; i != tmpS.length(); ++i) {
    
      
 if (tmpS[i] != ' ' && (tmpS[i] < '1' || tmpS[i] > '9')) {
    
      
 tmp = tmpS[i];
    
      
 tmpVal = 0;
    
      
 }
    
      
 else if (tmpS[i] != ' ') {
    
      
 tmpVal = tmpVal * 10 + tmpS[i] - '0';
    
      
 }
    
      
 }
    
      
 switch (tmp) {
    
      
 case '+' : befVal += tmpVal * 9 + x; break;
    
      
 case '-' : befVal -= tmpVal * 9 + x; break;
    
      
 }
    
      
 tmpS += " "; tmpS += tc; break;
    
      
 }
    
      
 if(x == n) {
    
      
 if (befVal == 0)
    
      
 cout << tmpS << endl;
    
      
 return;
    
      
 }
    
      
 dfs(x + 1, 2, befVal, tmpS);
    
      
 dfs(x + 1, 0, befVal, tmpS);
    
      
 dfs(x + 1, 1, befVal, tmpS);
    
      
}
    
      
int main () {
    
      
 freopen("zerosum.in", "r", stdin); 
    
      
 freopen("zerosum.out", "w", stdout);
    
      
 scanf("%d", &n);
    
      
 string ans("1");
    
      
 dfs(2, 2, 1, ans);
    
      
 dfs(2, 0, 1, ans);
    
      
 dfs(2, 1, 1, ans);
    
      
 return 0; 
    
      
}
   
     

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