[AOJ]Lesson - ITP1 Topic # 8 Character 字符
A: Toggling Cases
Write a program which converts uppercase/lowercase letters to lowercase/uppercase for a given string.
Input
A string is given in a line.
Output
Print the converted string in a line. Note that you do not need to convert any characters other than alphabetical letters.
Constraints
The length of the input string < 1200
Sample Input
fAIR, LATER, OCCASIONALLY CLOUDY.
Sample Output
Fair, later, occasionally cloudy.
题目大意
输入一个字符串，将大写字母变成小写，小写变成大写，并输出变化后的字符串。字符串长度小于1200。
#include <bits/stdc++.h>
using namespace std;
int main(){
string s;
getline(cin, s);
int len = s.size();
for(int i=0; i<len; i++){
if(s[i]>='A' && s[i]<='Z') // 大写变小写
s[i] += -'A'+'a';
else if(s[i]>='a' && s[i]<='z') // 小写变大写，注意有else
s[i] += -'a'+'A';
cout << s[i];
}
cout << endl;
return 0;
}
B: Sum of Numbers
Write a program which reads an integer and prints sum of its digits.
Input
The input consists of multiple datasets. For each dataset, an integer x is given in a line. The number of digits in x does not exceed 1000.
The input ends with a line including single zero. Your program should not process for this terminal symbol.
Output
For each dataset, print the sum of digits in x.
Sample Input
123
55
1000
0
Sample Output
6
10
1
题目大意
输入若干个数字字符串，求每个字符串中的数字之和。每个字符串长度不超过1000，最后一个字符串以“0”结束。
#include <bits/stdc++.h>
using namespace std;
int main(){
string s;
while(cin >> s){
if(s=="0")
break;
int ans = 0;
for(int i=0; i<s.size(); i++){
if(s[i]>='0' && s[i]<='9')
ans += s[i]-'0';
}
cout << ans << endl;
}
return 0;
}
C: Counting Characters
Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters.
Input
A sentence in English is given in several lines.
Output
Prints the number of alphabetical letters in the following format:
a : The number of 'a'
b : The number of 'b'
c : The number of 'c'
.
.
z : The number of 'z'
Constraints
The number of characters in the sentence < 1200
Sample Input
This is a pen.
Sample Output
a : 1
b : 0
c : 0
d : 0
e : 1
f : 0
g : 0
h : 1
i : 2
j : 0
k : 0
l : 0
m : 0
n : 1
o : 0
p : 1
q : 0
r : 0
s : 2
t : 1
u : 0
v : 0
w : 0
x : 0
y : 0
z : 0
题目大意
有一个若干行的英文句子，请统计字母a~z的个数。忽略大小写，句子长度不超过1200。
#include <bits/stdc++.h>
using namespace std;
int main(){
string s;
int a[26];
memset(a, 0, sizeof(a));
while(cin >> s){ // 若干行
int len = s.size();
for(int i=0; i<len; i++){
if(s[i]>='A' && s[i]<='Z') // 统一为小写
s[i] += 'a'-'A';
int t = s[i]-'a';
a[t] ++; // 统计数量
}
}
for(int i=0; i<26; i++){ // 输出
cout << (char) (i+'a') << " : " << a[i] << endl;
}
return 0;
}
D: Ring
Write a program which finds a pattern p in a ring shaped text s.
Input
In the first line, the text s is given.
In the second line, the pattern p is given.
Output
If p is in s, print Yes in a line, otherwise No.
Constraints
1≤ length of p≤ length of s≤100
s and p consists of lower-case letters
Sample Input 1
vanceknowledgetoad
advance
Sample Output 1
Yes
Sample Input 2
vanceknowledgetoad
advanced
Sample Output 2
No
题目大意
输入两个字符串s和p，第一个字符串s是一个环，问s中是否包含字符串p。如果包含，输出Yes,否则输出No。所有字母都是小写， p和s的长度大于0且小于等于100，并且p不会比s更长。
// 考察环的处理，以及查找函数find（)。
#include <bits/stdc++.h>
using namespace std;
int main(){
string s1, s2;
cin >> s1 >> s2;
s1 = s1+s1; // 环的处理，拼起来
if(s1.find(s2, 0)!=s1.npos) // s1.npos表示在s1中没有位置，即找不到
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}
