1117 数字之王

solution

• 判断现有数字是否全为个位数
  • 全为个位数，找出出现次数最多的数字，并首行输出最多出现次数，第二行输出所有出现该次数的数值
  • 不全为个位数
    • 若当前位数值为0，无需处理
    • 若当前位数值非0，则每位立方相乘，并把结果个位和相加更新为当前位新数值

#include<iostream>
using namespace std;
int n1, n2, flag[10] = {0}, a[1001], maxn = -1, judge = 0, first = 1, d;
int main(){scanf("%d%d", &n1, &n2);for(int i = n1; i <= n2; i++){a[i] = i;}if(n2 > 9) judge = 1;while(judge){judge = 0;for(int i = n1; i <= n2; i++){int t1 = 1, t2 = 0, t = a[i];if(t == 0) continue;while(t){d = t % 10;t /= 10;t1 *= d * d * d;}while(t1){d = t1 % 10;t1 /= 10;t2 += d;}a[i] = t2;if(t2 > 9) judge = 1;}}for(int i = n1; i <= n2; i++){flag[a[i]]++;}for(int i = 0; i < 10; i++){if(flag[i] > maxn) maxn = flag[i];}printf("%d\n", maxn);for(int i = 0; i < 10; i++){if(flag[i] == maxn){if(first) first = 0;else printf(" ");printf("%d", i);}}return 0;
} 

or

#include<iostream>
using namespace std;
int n1, n2, flag[10] = {0}, a[1001], maxn = -1, judge = 0, first = 1, d;
int main(){scanf("%d%d", &n1, &n2);for(int i = n1; i <= n2; i++) a[i] = i;if(n2 > 9) judge = 1;while(judge){judge = 0;for(int i = n1; i <= n2; i++){int t1 = 1, t2 = 0, t = a[i];if(t == 0) continue;while(t){d = t % 10;t /= 10;t1 *= d * d * d;}while(t1){d = t1 % 10;t1 /= 10;t2 += d;}a[i] = t2;if(t2 > 9) judge = 1;}}for(int i = n1; i <= n2; i++){flag[a[i]]++;if(flag[a[i]] > maxn) maxn = flag[a[i]];}printf("%d\n", maxn);for(int i = 0; i < 10; i++){if(flag[i] == maxn){if(first) first = 0;else printf(" ");printf("%d", i);}}return 0;
}