Question link: https://vjudge.net/problem/HDU-2266
Original title
How Many Equations Can You Find Problem Description
Now give you an string which only contains 0, 1 ,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9.You are asked to add the sign ‘+’ or ’-’ between the characters. Just like give you a string “12345”, you can work out a string “123+4-5”. Now give you an integer N, please tell me how many ways can you find to make the result of the string equal to N .You can only choose at most one sign between two adjacent characters.
input
Each case contains a string s and a number N . You may be sure the length of the string will not exceed 12 and the absolute value of N will not exceed 999999999999.
output
The output contains one line for each data set : the number of ways you can find to make the equation.
Sample Input
123456789 3
21 1Sample Output
18
1
Question meaning analysis:
The meaning of the question is to give a string s composed of numeric characters, and an integer N. You can add a '+' sign or a '-' sign to any two characters in s to make the value of the constructed mathematical formula equal to N. Ask how many different mathematical formulas there are for given s and N.
Idea:
Because the length of the string is short, just use ternary to enumerate the status of the possible operators at each position
. It is worth noting that the values of the numbers corresponding to all substrings of the string s need to be stored in the two-dimensional array to in advance. in, otherwise Time Limit will be exceeded.
Code:
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <string>
#define MAXN 100
using namespace std;
int main(){
int i, j, x, ans;
string s;
while (cin >> s >> ans){
int end = 1, k = s.length()-1, cnt = 0;
//k个运算数（每个运算数三种状态为+，-，或无）
for (i = 0; i < k; i++) end *= 3;
//to[i][j]表示字符串s中下标i开始，长度为j的数字大小
int to[MAXN][MAXN] = {
0};
for (i = 0; i < s.length(); i++){
for (j = 1; j < s.length()-i+1; j++){
to[i][j] = atoi(s.substr(i, j).data());
}
}
//枚举
for (i = 0; i < end; i++){
int a[MAXN];
//将j转换为k位3进制数
for (j = 0, x = i; j < k; j++){
a[j] = x%3;
x /= 3;
}
//temp储存分割所得数字，cal储存运算符状态
int temp[MAXN], nt = 0;
int cal[MAXN], nc = 0;
int pre = 0;
for (j = 0; j < k ;j++){
//如果a[j]不为0，代表运算符为+或-
if (a[j]){
// temp[nt++] = atoi(s.substr(pre, j+1-pre).data());
temp[nt++] = to[pre][j+1-pre];
pre = j+1;
cal[nc++] = a[j];
}
}
// temp[nt++] = atoi(s.substr(pre, j+1-pre).data());
temp[nt++] = to[pre][j+1-pre];
int sum = temp[0];
for (j = 1; j < nt; j++){
if (cal[j-1] == 1){
sum += temp[j];
}else{
sum -= temp[j];
}
}
cnt += (sum == ans);
}
cout << cnt << endl;
}
return 0;
}