1023 Have Fun with Numbers 大数 哈希？？？

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 different 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 kdigits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input 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
第一次没看到数据规定 然后写了个溢出的程序

#include<iostream>
#include<cstdio>
#include<cmath>
#define ll long long
using namespace std;
int main()
{
    ll n,a[10]={0},m,r,k;
    bool flag=true;
    cin>>n;
    m=n*2;
    k=m;
    while(n>9)
    {
      r=n%10;
      n=n/10;
      a[r]++;;
    }
    a[n]++;
    while(m>9)
    {
      r=m%10;
      m=m/10;
      a[r]--;
    }
    a[m]--;
    for(int i=0;i<10;i++)
    {
        if(a[i]!=0)
        flag=false;
    }
    if(flag==true)
    cout<<"Yes"<<endl;
    else
    cout<<"No"<<endl;
    cout<<k<<endl;
    return 0;
}

看了别人的才知道使用string  哭  算第一次做这种题吧

unsigned int 0～4294967295
int 2147483648～2147483647
unsigned long 0～4294967295
long 2147483648～2147483647
long long的最大值：9223372036854775807 （刚好19位）
long long的最小值：-9223372036854775808
unsigned long long的最大值：18446744073709551615
__int64的最大值：9223372036854775807
__int64的最小值：-9223372036854775808
unsigned __int64的最大值：18446744073709551615
-柳神代码

#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
" : "Yes
");
    if(flag == 1) printf("1");
    printf("%s", num);
    return 0;
}