According to Bartjens UVA - 817

题目传送门

题意：输入一个以等号结尾。前面只包含数字的表达式，插入一些加号、减号和乘号，使得运算结果等于2000.表达式里的整数不能有前导零（例如，0100和000都是非法的），运算符都是二元的（例如，2*-100*-10+0=是非法的），并且符合通常的运算优先级法则。输入数字个数不超过9。如果有多解任意顺序输出：如果无解，输出IMPOSSIBLE。例如，2100100=有3组解，按照字典序依次为2*100*10+0=、2*100*10-0=和2100-100=。

思路：枚举所有的情况，去除掉非法的情况然后进行计算就可以了。要注意一定要往里面插入符号所以2000=应该是输出IMPOSSIBLE，但是如果是02000=就可以输出0+2000=（一直错在这个地方）。

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>

#define MAXN 1010
#define MAXE 210
#define INF 10000000
#define MOD 1000000007
#define LL long long
#define pi acos(-1.0)

using namespace std;

int len;
vector<string> vec;
set<string> Set;
string str;
string op = "*+-";

void dfs(int cur, string temp) {
if (cur == len - 1) {
vec.push_back(temp);
return;
}
if (temp[temp.length() - 1] == '0') {
if (temp.length() >= 2 && isdigit(temp[temp.length() - 2])) {
dfs(cur + 1, temp + str[cur + 1]);
}
for (int i = 0; i < 3; ++i) {
dfs(cur + 1, temp + op[i] + str[cur + 1]);
}
} else {
for (int i = 0; i < 3; ++i) {
dfs(cur + 1, temp + op[i] + str[cur + 1]);
}
dfs(cur + 1, temp + str[cur + 1]);
}
}

void calculate(string temp) {
vector<LL> Num, Sign;
for (int i = 0; i < temp.size(); ++i) {
LL ans = 0;
while (i < temp.length() && isdigit(temp[i])) {
ans = ans * 10 + temp[i] - '0';
i++;
}
Num.push_back(ans);
if (i == temp.length())
break;
Sign.push_back(temp[i]);
}
for (int i = 0; i < Sign.size(); i++)
if (Sign[i] == '*') {
Num[i + 1] *= Num[i];
Num.erase(Num.begin() + i);
Sign.erase(Sign.begin() + i);
i--;
}

LL sum = Num[0];
for (int i = 0; i < Sign.size(); i++)
if (Sign[i] == '+')
sum += Num[i + 1];
else
sum -= Num[i + 1];
if (sum == 2000) {
Set.insert(temp);
}
}

int main() {
std::ios::sync_with_stdio(false);
int kase = 0;
while (cin >> str && str != "=") {
vec.clear();
Set.clear();
LL ans = 0;
for (int i = 0; i < str.length() - 1; ++i) {
ans = ans * 10 + str[i] - '0';
}
len = (int)str.length() - 1;
str = str.substr(0, len);
cout << "Problem " << ++kase << endl;
if (str == "2000") {
cout << "  IMPOSSIBLE\n";
} else {
string temp = "\0";
temp += str[0];
dfs(0, temp);
for (int i = 0; i < vec.size(); ++i) {
calculate(vec[i]);
}
if (Set.size()) {
for (set<string>::iterator it = Set.begin(); it != Set.end(); ++it) {
cout << "  ";
cout << *it << "=\n";
}
} else {
cout << "  IMPOSSIBLE\n";
}
}
}
return 0;
}

/*
2100100=
77=
=
*/