7-1 Forever (20 分)
“Forever number” is a positive integer A with K digits, satisfying the following constrains:
the sum of all the digits of A is m;
the sum of all the digits of A+1 is n; and
the greatest common divisor of m and n is a prime number which is greater than 2.
Now you are supposed to find these forever numbers.
Input Specification:
Each input file contains one test case. For each test case, the first line contains a positive integer N (≤5). Then N lines follow, each gives a pair of K (3<K<10) and m (1<m<90), of which the meanings are given in the problem description.
Output Specification:
For each pair of K and m, first print in a line Case X, where X is the case index (starts from 1). Then print n and A in the following line. The numbers must be separated by a space. If the solution is not unique, output in the ascending order of n. If still not unique, output in the ascending order of A. If there is no solution, output No Solution.
测试用例
input:2645780
output:
Case 1101899991027999910369999104599991054999910639999107299991081999910909999
Case 2
No Solution
待验证代码
是否能成功ac,待验证
#include<iostream>#include<cmath>
using namespace std;
bool isprime(int a){if(a<=2)return false;for(int i=2; i*i<=a; i++)if(a%i==0)return false;return true;}intgcd(int a,int b){return b==0? a :gcd(b, a%b);}intsums(int a){int sum =0;while(a!=0){
sum +=(a%10);
a/=10;}return sum;}intmain(){int n, k, m;
cin >> n;for(int i=0; i<n; i++){scanf("%d %d",&k,&m);printf("Case %d\n", i+1);int left =pow(10,k-3), right =pow(10,k-2)-1, flag =0;for(int i=left; i<=right; i++){int mm =sums(i*100+99), mn =sums(i*100+100);if(mm==m &&isprime(gcd(mm,mn))){printf("%d %d\n", mn, i*100+99);
flag =1;}}if(!flag)printf("No Solution\n");}return0;}
原因分析
前置条件
K (3<K<10)
m (1<m<90)
the sum of all the digits of A is m;
the sum of all the digits of A+1 is n;
the greatest common divisor of m and n is a prime number which is greater than 2.