UVA10317-Equating Equations (backtracking + pruning)

UVA10317-Equating Equations (backtracking + pruning)

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Question: Give a formula, but this formula is not necessarily an equation. In the case of '+', '-', '=', the positions of digits remain unchanged, make it an equation and, if possible, output an arrangement.

Idea: we move all the numbers on the right of the equal sign to the right of the equal sign, for example, a + B-c = d-e. After moving it to a + B + e-(c + d) = 0, that is, a + B + e = c + d. So when the formula sub can change to an equation, the sum of all numbers must be an even number. Then the problem can be converted to finding the number of m in the n number (the value of m is the number of integers on the left of the equal sign), so that the sum of m numbers is half of the sum of sums.

#include 
 
  #include 
  
   #include 
   
    #include using namespace std;const int MAXN = 1005;char str[MAXN], Lsy[MAXN], Rsy[MAXN], vis[MAXN];int arr[MAXN], front[MAXN], back[MAXN];int cnt1, cnt2, eql, Lnum, ans, flag, L, R;int init() {    cnt1 = 1, cnt2 = eql = L = R = 0;    int l = strlen(str), sum = 0;     sscanf(str, "%d", &arr[0]);    sum = arr[0];    for (int i = 0; i < l; i++) {        if (str[i] == '+' || str[i] == '-' || str[i] == '=') {            if (str[i] == '=')                eql = i;            if (!eql) {                Lsy[L] = str[i];                L++;            }            else if (eql != i) {                Rsy[R] = str[i];                R++;            }            sscanf(str + i + 1, "%d", &arr[cnt1]);            sum += arr[cnt1++];        }     }    Lnum = 1;    for (int i = 0; i < l; i++) {        if (i < eql && str[i] == '+')             Lnum++;        else if (i > eql && str[i] == '-')            Lnum++;     }    return sum;}int dfs(int k, int pos, int cur) {    if (k == Lnum) {        if (cur == ans)              return true;        return false;    }    if (Lnum - k > cnt1 - pos)        return false;    if (pos < cnt1 && cur + arr[pos] <= ans) {        vis[pos] = 1;         if (dfs(k + 1, pos + 1, cur + arr[pos]))             return true;        vis[pos] = 0;    }    if (pos < cnt1 && dfs(k, pos + 1, cur))         return true;    return false;}void outPut() {    int x = 0, y = 0;    for (int i = 0; i < cnt1; i++) {        if (vis[i])             front[x++] = arr[i];        else            back[y++] = arr[i];     }    printf("%d", front[--x]);    for (int i = 0; i < L; i++) {        printf(" %c ", Lsy[i]);         if (Lsy[i] == '+')            printf("%d", front[--x]);        if (Lsy[i] == '-')            printf("%d", back[--y]);     }    printf(" = ");    printf("%d", back[--y]);     for (int i = 0; i < R; i++) {        printf(" %c ", Rsy[i]);         if (Rsy[i] == '+')            printf("%d", back[--y]);          if (Rsy[i] == '-')            printf("%d", front[--x]);    }    printf("\n");}int main() {    while (gets(str)) {        int s = init();         if (s % 2)            printf("no solution\n");        else {            ans = s / 2;            memset(vis, 0, sizeof(vis));            if (dfs(0, 0, 0))                outPut();              else                 printf("no solution\n");        }    }        return 0;}