PAT 1023 Have Fun with Numbers

#include <cstdio>

#include <cstdlib>

#include <vector>



using namespace std;



bool double_num(vector<char> &num) {

    int carry = 0;

    int i = num.size() - 1;

    while (i>=0) {

        int d = num[i] * 2 + carry;

        carry = d / 10;

        num[i]= d % 10;

        i--;

    }

    return carry < 1;

}



void stat_count(vector<char>& num, int* cnt) {

    for (int i=num.size() - 1; i>=0; i--) {

        cnt[num[i]]++;

    }

}



bool is_same_digits(vector<char>& num, int* cnt) {

    int cur_cnt[10] = {0};



    stat_count(num, cur_cnt);

    

    for (int i=0; i<10; i++) {

        if (cnt[i] != cur_cnt[i]) return false;

    }

    return true;

}



void print(vector<char> &num) {

    int len = num.size();

    for (int i=0; i<len; i++) {

        printf("%d", num[i]);

    }

    printf("\n");

}



int main() {

    const int MAX_DIGITS = 20;



    vector<char> num;



    char buf[MAX_DIGITS + 1];



    int count[10] = {0};

    

    scanf("%s", buf);

    

    for (int i=0; buf[i] != '\0'; i++) {

        if (num.empty() && buf[i] == '0') continue; // skip leading zeros

        num.push_back(buf[i] - '0');

    }

    if (num.empty()) num.push_back(0);

    

    // stat original number digit count

    stat_count(num, count);

    

    bool found = false;

    bool no_carry = double_num(num);

    

    bool is_same = is_same_digits(num, count);

    if (is_same) {

        printf("Yes\n");

    } else {

        printf("No\n");

    }

    if (!no_carry) printf("1");

    print(num);

    return 0;

}

我觉得玩这种输出陷阱很没意思，给个反向例子会死么