1023 Have Fun with Numbers (20 分)
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 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 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:
代码语言：javascript复制1234567899Sample Output:
代码语言：javascript复制Yes
2469135798思路：本是一道水的不行的20分题 ，就随便写了一个交了发现一半的测试点错了，后来看数据范围才知道，最多不超过20位，
emmm，longlong可能有些数据不行，
首先我们复习一下整型、长整型的数据表示范围
unsigned int 0～4294967295 int -2147483648～2147483647 unsigned long 0～4294967295 long -2147483648～2147483647 long long的最大值：9223372036854775807 long long的最小值：-9223372036854775808 unsigned long long的最大值：1844674407370955161
__int64的最大值：9223372036854775807 __int64的最小值：-9223372036854775808 unsigned __int64的最大值：18446744073709551615
很明显我们可以看出哪怕是unsigned long long也只有20位，首位还是1，如果极端数据是99999999999999999999呢，那这个时候我们就要用字符串了
这道题目其实是一个很弱的大数相乘，只乘2，如果得数位数和原数不一样的一律输出No，道理很简单，位数都不相同，那组成肯定也不一样~，大体就是写了个乘法函数，然后去求两个数的组成，一样就Yes，否则No
注意输出乘2的得数有可能有进位！
希望对您有所帮助~
代码语言：javascript复制// luogu-judger-enable-o2
#include<bits/stdc .h>
#include<unordered_set>
#define rg register ll
#define inf 2147483647
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
#define ll long long
#define maxn 100005
#define lb(x) (x&(-x))
const double eps = 1e-6;
using namespace std;
inline ll read()
{
char ch = getchar(); ll s = 0, w = 1;
while (ch < 48 || ch>57) { if (ch == '-')w = -1; ch = getchar(); }
while (ch >= 48 && ch <= 57) { s = (s << 1) (s << 3) (ch ^ 48); ch = getchar(); }
return s * w;
}
inline void write(ll x)
{
if (x < 0)putchar('-'), x = -x;
if (x > 9)write(x / 10);
putchar(x % 10 48);
}
char a[25],b[25];
ll c[10],d[10];
int main()
{
cin>>a;
for(rg i=0;a[i];i )
{
b[strlen(a)-1-i]=a[i];
}
ll temp=0,tot=strlen(b);
for(rg i=0;b[i];i )
{
ll tep=(b[i]-48)*2 temp;
temp=tep/10;
tep%=10;
b[i]=tep 48;
}
while(temp)
{
b[tot ]=temp 48;
temp/=10;
}
ll sum=0;
for(rg i=0;i<strlen(a);i )
{
c[a[i]-48] ,d[b[i]-48] ;
}
for(rg i=0;i<10;i )
{
sum =(c[i]==d[i]);
}
if(strlen(b)!=strlen(a))
{
cout<<"No"<<endl;
for(rg i=strlen(b)-1;b[i];i--)cout<<b[i];
return 0;
}
sum==10?cout<<"Yes"<<endl:cout<<"No"<<endl;
for(rg i=strlen(b)-1;b[i];i--)cout<<b[i];
return 0;
}
