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;
}