uva 817According to Bartjens

说到这题 我只想说，uva的测试数据真牛逼，面面俱到

此题其实就是一个暴力搜索，尝试完所有的情况，看其是否等于2000就行了，

一下是关键的几点 ：2000=是 不可能的；注意排除具有前导0的数字的情况，

////  main.cpp//  uva 817 - According to Bartjens////  Created by XD on 15/8/3.//  Copyright (c) 2015年 XD. All rights reserved.////暴力搜索#include <iostream>#include <string>#include <queue>#include <stack>#include <stdio.h>#include <stdlib.h>#include <math.h>#include<vector>#include <string.h>#include <string>#include <algorithm>#include <set>#include <map>#include <cstdio>using namespace std ;char inter = 'A'  ;//定义操作符char operation[3] = {'+','-' ,'*'} ;char s[30] ;int n  ;  //有多少个位置可以填充int md ; //有多少个位置插入了符号set<string> ans ;   //存放结果的集合vector<int > pos ;   //存放当前情况下的填充位置bool isnumber(char ch){    return ch >= '0' && ch <= '9' ;}//获取数字---如果是*连接的要在这里算出来-知道遇到+或者减号int getnum(int *p){    char num[10] ;    int j = 0 ;    while (*p  < 2 * n + 1 && s[*p]!='+'&&s[*p]!= '-' ) {        if (isnumber(s[*p])) {            num[j++] = s[*p] ;        }        else{            if (s[*p] == '*') {                if (num[0] == '0' && j > 1) {                    return -1 ;                }                else{                    num[j] = '\0' ;                    ++(*p) ;                    int t = getnum(p) ;                    if (t!=-1) {                        return atoi(num) *t  ;                    }                    return -1 ;                }            }        }        (*p)++ ;    }    num[j] = '\0' ;    if (num[0] == '0' && j > 1) {        return -1 ;    }        return atoi(num) ;}//运算函数int run(int a , int b  , char operation){    switch (operation) {        case '+':            return a + b ;            break;        case '-':            return a - b ;        default:            return a*b ;            break;    }}//判断是否等于2000bool equal2000(){    int a =0 ,b,ope = '+'  ;    for(int i = 0 ; i < 2 * n  + 1 ; i++)  {        if(isnumber(s[i])){            b =getnum(&i) ;            if(b==-1)            {                return false ;            }            i-- ;            a = run(a , b , ope) ;        }        else if (s[i] == '+' || s[i] == '-')        {            ope = s[i] ;        }    }    return a == 2000 ? true:false  ;}void  dfs(int d){    if (d == md) {        if (equal2000()) {            string result  ;            for (int  i = 0 ; i < 2*n + 1; i++) {                if(isnumber(s[i]) || s[i] == '*'|| s[i] == '+'||s[i] == '-')                {                    result.operator+=(s[i]) ;                }            }                        ans.insert(result) ;        }    }    else    {        for (int i = 0; i < 3; i++) {            s[pos[d] * 2  + 1] = operation[i] ;            dfs(d + 1) ;        }    }}void solved(){    int end = 1 << n ;    for(int i = 1 ; i <end ; i++ )    {        pos.clear() ;        for (int j = 0; j < n; j++) {            if (i &(1 << j)) {                pos.push_back(j) ;            }        }        md = (int )pos.size()  ;        dfs(0) ;//        int len =(int ) pos.size() ;        for (int j =0 ; j < md; j++) {            s[2*pos[j]+1] = inter ;        }    }}int main() {    char temp[15] ;    int casenum =0 ;    while (scanf(" %s" , temp)==1 && temp[0]!='=') {        int len = (int )strlen(temp) ;        s[len-1] ='\0' ;        for (int i = 0; i < len-1; i++) {            s[i*2] = temp[i] ; s[i*2+1] = inter ;        }        s[2*(len-2)+1] = '\0' ;        n = len  -2 ;        ans.clear() ;        solved() ;        len = (int )ans.size() ;        cout <<"Problem "<<++casenum <<endl;        if (len == 0 ) {            printf("  IMPOSSIBLE\n") ;        }        else{            for (set<string> ::iterator it = ans.begin(); it != ans.end(); it++) {                cout <<"  "<<*it <<"="<<endl ;            }        }    }        return 0;}