USACO 2.3.3 deep search for Zero Sum

Zero SumConsider the sequence of digits from 1 through N (where N = 9) in increasing order: 1 2 3... n. now insert either a' + 'for addition or a'-' for subtraction or A' [blank] to run the digits together between each pair of digits (not in front of the first digit ). calculate the result that of the expression and see if you get zero. write a program that will find all sequences of length N that Produce a zero sum. program name: zerosumINPUT FORMATA single line with the integer N (3 <= N <= 9 ). sample input (file zerosum. in) 7 OUTPUT FORMATIn ASCII order, show each sequence that can create 0 sum with a '+', '-', or ''between each pair of numbers. sample output (file zerosum. out) 1 + 2-3 + 4-5-6 + 71 + 2-3-4 + 5 + 6-71-2 3 + 4 + 5 + 6 + 71-2 3- 4 5 + 6 71-2 + 3 + 4-5 + 6-71-2-3-4-5 + 6 + 7 /*----------------------------------- -------------------------- */Once again, it indicates that USACO will help you review the question. This question is not a dynamic conversion, but a simple deep search. Simply traverse all the situations. A maximum of 9 numbers, 8 symbols, and a maximum of 3 8 power types can be obtained. Program: [cpp]/* ID: zhaorui3 PROG: zerosum LANG: C ++ */# include <iostream> # include <fstream> using namespace std; int operations [9] = {0}; // 0: nothing, 1: +, 2:-int templeft = 0, tempRight = 0; int main () {operations [0] = 1; ifstream fin ("zerosum. in "); ofstream fout (" zerosum. out "); int end; fin> end; while (true) // this layer of loop is used to traverse all symbols {templeft = 0; tempRight = 0; for (int j = 1; j <= end; j ++) // this layer of loop is used to calculate the result under the current symbol {if (operations [J-1] = 1) // + {int k = j; tempRight = j; while (k <= end-1 & operations [k] = 0) {tempRight = tempRight * 10 + k + 1; k ++; j ++;} templeft = templeft + tempRight; tempRight = 0; continue ;} if (operations [J-1] = 2) //-{int k = j; tempRight = j; while (k <= end-1 & operations [k] = 0) {tempRight = tempRight * 10 + k + 1; k ++; j ++ ;} templeft = templeft-tempRight; tempRight = 0; continue ;}}if (templeft = 0) {fout <1; for (int m = 0; m <= end-2; m ++) {switch (operations [m + 1]) {case 1: fout <'+'; break; case 2: fout <'-'; break; case 0: fout <''; break;} fout <m + 2;} fout <endl ;} operations [end-1] ++; for (int m = end-1; m> 1; m --) {if (operations [m]> 2) {operations [m] = 0; operations [M-1] ++ ;}} if (operations [1]> 2) break ;}}