标签：rom   nal   lse   ack   字符串转换   string   which   ==   input   

题目

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a diferent permutation. Check to see the result if we double it
again!
Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.
Input Specification:
Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.
Output Specification:
For each test case, first print in a line “Yes” if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or “No” if not. Then in the next line, print the doubled number.
Sample Input:
1234567899
Sample Output:
Yes
2469135798

题目分析

已知一个最多20位的数字N，判断N乘2后的数字M是否只是组成N的各位数字的重新排列

解题思路

1. 用一个数组记录N中各位数字出现的次数，乘2后的数字M中出现相同数字则出现次数减一（该判断隐含M和N的长度不相等的情况，长度不相等则出现次数一定不相等，数组中出现次数不相等时元素值不为0）
2. 大整数相乘算法

易错点

1. 字符和数字的转换（字符-‘0‘=数字）

知识点

1. 字符串转换

#include <algorithm>
reverse(ds.begin(),ds.end());

Code

Code 01

#include <iostream>
#include <algorithm>
using namespace std;
int main(int argc,char * argv[]) {
    string s,ds;
    cin>>s;
    int sf[10]= {0},carry=0;
    for(int i=s.length()-1; i>=0; i--) {
        sf[s[i]-'0']++;
        int temp=(s[i]-'0')*2+carry;
        ds.push_back(temp%10+'0');
        carry=temp/10;
        sf[temp%10]--;
    }
    while(carry!=0) {
        ds.push_back(carry%10+'0');
        sf[carry%10]--;
        carry/=10;
    }

    bool flag = true;

//  strrev(s.c_str());
//  if(ds.length()!=s.length())flag=false;
//  else {}
    for(int i=0; i<10; i++) {
        if(sf[i]!=0) {
            flag = false;
            break;
        }
    }
    printf("%s", flag? "Yes\n" : "No\n");
    reverse(ds.begin(),ds.end());
    printf("%s",ds.c_str());
    return 0;
}

Code 02

#include <cstdio>
#include <string.h>
using namespace std;
int book[10];
int main() {
    char num[22];
    scanf("%s", num);
    int flag = 0, len = strlen(num);
    for(int i = len - 1; i >= 0; i--) {
        int temp = num[i] - '0';
        book[temp]++;
        temp = temp * 2 + flag;
        flag = 0;
        if(temp >= 10) {
            temp = temp - 10;
            flag = 1;
        }
        num[i] = (temp + '0');
        book[temp]--;
    }
    int flag1 = 0;
    for(int i = 0; i < 10; i++) {
        if(book[i] != 0)
            flag1 = 1;
    }
    printf("%s", (flag == 1 || flag1 == 1) ? "No\n" : "Yes\n");
    if(flag == 1) printf("1");
    printf("%s", num);
    return 0;
}

PAT Advanced 1023 Have Fun with Numbers (20) [?整数运算]

标签：rom   nal   lse   ack   字符串转换   string   which   ==   input   

原文地址：https://www.cnblogs.com/houzm/p/12268856.html