题目链接:http://www.patest.cn/contests/pat-a-practise/1023
题目:
1023. Have Fun with Numbers (20)
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 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:1234567899Sample Output:
Yes 2469135798
分析:
给你一个数(大数),判断其*2的结果数是否和原数是一样的排列,注意这里不是说给出的数都是1-9的排列,只是题目中举例子用到了1-9而已。
这里设置了ans[]数组,对于原数的每个数都++,对于结果数的每个数都--,那么最后只要判断ans是否全都0就可以判断原数和结果数是否是相同的排列
注意:
考虑到进位,乘以2以后可能会多出一位,而且20位的数字要用string表示而不是用long long表示。
可以看到以下数字的表示范围中long long不够20位。
int,long,long long类型的范围
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
AC代码:
#include<stdio.h> using namespace std; int ans[10]; char num1[22]; char num2[22]; int main(void){ //freopen("F://Temp/input.txt","r",stdin); while (scanf("%s",num1) != EOF){ for (int k = 0; k< 10; k++){ ans[k] = 0; } int di = 0,jin = 0,ji = 0; int i; for (i = 21; num1[i] == 0; i--);//找到最后一位的下标开始计算 for ( ; i >= 0; i -- ){ ji = (num1[i] - '0') * 2; ans[num1[i] - '0'] ++;//ans对原数相应位的个数++ di = ji % 10;//*2后的当前位的数字 num2[i] = di + jin + '0'; ans[num2[i] - '0'] --;//ans对结果数的相应位的个数-- jin = (ji + jin) / 10; } if (jin != 0)ans[jin] ++; for (i = 1; i < 10; i++){ if (ans[i] != 0)break; }//判断ans是否全部都为0,若是,则说明原数和结果数是相同的排列 if (i == 10){ puts("Yes"); } else { puts("No"); } if (jin != 0)printf("%d",jin); puts(num2); } return 0; }
截图:
——Apie陈小旭