int numberOfOnes(const char* str)
{
if(str == NULL || *str < '0' || *str > '9' || *str == '\0')
{
return 0;
}
int first = *str - '0';
unsigned int length = static_cast<unsigned int>(strlen(str));
if(length == 1 && first == 0)
{
return 0;
}
if(length == 1 && first > 0)
{
return 1;
}
//假设str是"21345", numFirstDigit是10000~19999中1的数目
int numFirstDigit = 0;
if(first > 1)
{
numFirstDigit = powerBase10(length - 1);
}
else if(first == 1)
{
numFirstDigit = atoi(str + 1) + 1;
}
//numOtherDigit是数字10000~19999的第一位之外的数位中的数目
int numOtherDigit = first * (length - 1) * powerBase10(length - 2);
int numRecursive = numberOfOnes(str + 1);
return numFirstDigit + numOtherDigit + numRecursive;
}
//方法二
int findOnesFromOneToN_Solution2(int n)
{
if(n <= 0)
{
return 0;
}
char str[50];
sprintf(str, "%d", n);
return numberOfOnes(str);
}
int main() {
int n,m;
scanf("%d %d",&n, &m);
long long int * skills = malloc(n*(sizeof(long long int)));
long long parsed;
int i,j;
char str[500][500];
for(i=0;i<n;i++){
char * ptr;
scanf("%s",str[i]);
//parsed = strtoll(str, & ptr, 2);
// printf("%llX\n", parsed);
//skills[i] = parsed;
}
// qsort(skills,n,sizeof(skills[0]),comp);
long long maxteam=0;
long long maxskill=0;
long long temp;
for(i=0;i<n;i++){
for(j=i+1;j<n;j++){
temp = numberOfOnes(str[i],str[j],m);
if(temp>maxskill){
maxskill=temp;
maxteam =1;
}
else if(temp == maxskill){
maxteam++;
}
}
}
//maxskill = numberOfOnes(maxskill);
printf("%lld\n%lld\n",maxskill,maxteam);
return EXIT_SUCCESS;
}
