C语言（C班）B题题解
fgets()函数可以用来直接读取一行输入。参数：地址、数量、输入流（这里是标准输入stdin）
#include <stdio.h>
int main()
{
double p[26]={0}, max=0;
char word[4000];
int i, mi, cnt=0;
fgets(word, 4000, stdin);
char *a=word;
while(*a!='\n')
{
if(*a==' ') {a++; continue;}
else {p[*a-'a']++; cnt++;}
a++;
}
for(i=0; i<26; i++) if(max<p[i]) {max=p[i]; mi=i;}
printf("%c %.2lf\n",'a'+mi, max/cnt);
return 0;
}
def main():
str_input = input()
count = {}
str_not_split = str_input.replace(" ", "")
str_length = len(str_not_split)
for i in set(str_not_split):
count[i] = str_not_split.count(i)
count_result = sorted(count.items(), key=lambda x: (-x[1], x[0]))
max_value = count_result[0][1]
print('%c %.2f' % (count_result[0][0],max_value / str_length))
if __name__ == '__main__':
main()
会用STL就行了
#include <bits/stdc++.h>
#include <algorithm>
using namespace std;
map<char,int> mp;
set<char> se;
int main() {
string s;
getline(cin,s);
int cnt = 0;
for(int i = 0;i < s.length();i++) {
if(s[i]!=' '){
cnt++;
mp[s[i]]++;
}
}
int max = -1;
for(auto it = mp.begin();it != mp.end();it++) {
int count = it->second;
if(count > max) max = count;
}
for(auto it = mp.begin();it != mp.end();it++) {
if(it->second == max) {
se.insert(it->first);
}
}
auto it = se.begin();
printf("%c %.2lf",*it,(double)mp[*it]/cnt);
return 0;
}
一种粗鲁的写法
using namespace std;
int ci[300],tot=0;
int amax=0;
char maxch;
int main()
{
memset(ci,0,sizeof(ci));
char tmp;
while((tmp=getchar())!=EOF)
if(isalpha(tmp)) {ci[tmp]++;tot++;if(ci[tmp]>amax) {amax=ci[tmp];maxch=tmp;}}
printf(“%c %.2f\n”,maxch,1.0*amax/tot);
return 0;
}
如果次数数组不用拼音，只开26个元素，就很优雅了
#include <bits/stdc++.h>
using namespace std;
typedef struct alpha{char c; int t = 0; int sub;}alpha;
bool cmp(const alpha &x, const alpha &y)
{
if(x.t == y.t)
return x.sub < y.sub;
else
return x.t > y.t;
}
int main()
{
string s;
getline(cin, s);
vector <alpha> a(26);
int space = 0;
for(int i = 0; i < s.size(); i++)
{
if(s[i] == ' ')
{
space++;
continue;
}
a[s[i] - 'a'].sub = s[i] - 'a';
a[s[i] - 'a'].t++;
}
sort(a.begin(), a.end(), cmp);
int len = s.size() - space;
double ans = a[0].t * 1.0 / len;
printf("%c %.2lf\n", a[0].sub + 'a', ans);
return 0;
}
不得不说，用python刷这种简单的实现题真的爽……
s = input()
counts = dict()
for c in [chr(i) for i in range(97,123)]:
counts[c] = s.count(c)
max_char = max(counts, key = counts.get)
print(‘%c %.2f’ % (max_char,counts[max_char]/(len(s) - s.count(‘ ‘))))
写的比较复杂，但是方法比较笨，当作多关键字排序写的
//2852 to count the frequency of a letter. multiple key element sorting
struct Word
{
char ch;
int times;
double possibility;
}word[26];
int cmp(const void a,const void b)
{
struct Word x=(struct Word) a;
struct Word y=(struct Word) b;
if(x->possibilitypossibility) return 1;
else if(x->possibility==y->possibility&&x->ch>y->ch) return 1;
else return -1;
}
int main()
{
char str[4000];
char text[3000];
gets(str);
int i,j=0;
int len1=strlen(str);
for(i=0;i<len1;i++)
if(str[i]!=’ ‘) text[j++]=str[i];//removal of space char
int len2=j;
for(i=0;i<26;i++){
word[i].times=1;
word[i].possibility=0.0;
}//structure initialization
int l=0;
for(i=0;i<len2;i++)
if(text[i]!='0'){
word[l].ch=text[i];
for(j=i+1;j<len2;j++)
if(text[j]==text[i]) {
word[l].times++;
text[j]='0';
}
l++;
}//input of key element
int len3=l;
for(i=0;i<len3;i++) word[i].possibility=(double)word[i].times/len2;
qsort(word,len3,sizeof(word[0]),cmp);
printf("%c %.2lf",word[0].ch,word[0].possibility);
return 0;
}
那个a!=’\n’为啥不能直接写a啊