Problem I. Hall of Fame
来源：ACM International Collegiate Programming Contest, Syrian Collegiate Programming Contest (2014) Syria, September, 23, 2014 CF链接Gym 100500I
ICPC World Finals 2014 was held in Russia, in Ekaterinburg from 21/06/2014 to 26/06/2014, hosted by Ural Federal University (UrFU). The opening ceremony was held on 23/06/2014 at 4:00 PM, in Cosmos Big Hall. During the ceremony a nice lady was singing the famous song "Hall of Fame". In case you don’t know the song here is its lyrics:（歌词在下图放大可查看）
Coach Fegla has invented a song effectiveness measurement methodology. This methodology in simple English means count the frequency of each character of the English letters [A-Z] in the given song (case insensitive). Get the top 5 characters with the highest frequencies, if 2 characters have the same frequency choose the one which is lexicographically larger. Sum up the indexes of these characters where the index of A is 0, index of B is 1, ..., index of of Z = 25. If this sum exceeds 62 print "Effective"otherwise print "Ineffective"without the quotes.
Input
The first line will be the number of test cases T. Each test case will consist of a series of words consisting only of upper or lower case English letters. Each test case ends with a line containing only "*"without the double quotes. Each word consists of a minimum of 1 letter and a max of 20 letters. The number of words in each test case will be a max of 20,000.
Output
For each test case print a single line containing: Case x: y x is the case number starting from 1. y is is the required answer either "Effective"or "Ineffective"without the quotes.
Examples
Fame.in：
2
You can be the greatest
*
You can be the best
You can be the King Kong banging on
your chest
*
Standard Output：
Case 1: Effective
Case 2: Ineffective
其实题目仔细分析就好，并不是很难，写代码的时候要细心。（题目意思就是：找出出现次数最多的那5个字母，如果出现次数相同则按字典序由大到小排序，A到Z分别指代0到25，最后将那5个字母代表的数字加起来，判断是否大于64，是则输出Effective，反之Ineffective。）
代码如下：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
struct node
{
int data, x; // data用来存某个字母出现的次数，x来存编号代表相应字母（0->A/a,1->B/b...）
} b[27];
void qs(int l, int r);
int main()
{
int T;
scanf("%d", &T);
int i = 0, g = 1;
while(T)
{
getchar();
char a[20020][22];
scanf("%s", a[++i]); // 每次输入一个小子段并存下来
if(strcmp(a[i], "*") == 0) // 每当遇到*则判断，if语句中最后重置i为0，并且T--
{
memset(b, 0, sizeof(b)); // 将结构体数组初始化
for(int j = 1; j < i; j++)
{
int l = strlen(a[j]);
for(int k = 0; k < l; k++) // 判断A/a...Z/z并存下来相应编号
{
if(a[j][k] >= 'A' && a[j][k] <= 'Z')
{
b[a[j][k] - 'A'].data++;
b[a[j][k] - 'A'].x = a[j][k] - 'A';
}
else if(a[j][k] >= 'a' && a[j][k] <= 'z')//因为不分大小写，所以A与a使用同一下标
{
b[a[j][k] - 'a'].data++;
b[a[j][k] - 'a'].x = a[j][k] - 'a';
}
}
}
qs(0, 25); // 快排
for(i = 0; i < 25; i++)
{
for(int j = 0; j < 25 - i; j++)
{
if(b[j].data == b[j + 1].data && b[j].x < b[j + 1].x)
// 因为想要在快排中出现次数相同时直接判断字典序，但是失败了，所以这里写了一个冒泡只把出现次数相同的排一下字典序，很幸运没有超时^.^
{
struct node t = b[j];
b[j] = b[j + 1];
b[j + 1] = t;
}
}
}
int sum = 0;
for(i = 0; i < 5; i++)
{
sum += b[i].x; // 一切准备就绪，只需要将前5个字母代表的数值加起来就好
}
if(sum > 62) printf("Case %d: Effective\n", g++);
else printf("Case %d: Ineffective\n", g++);
i = 0;
T--; // 不要忘了组数减减，以及上边重置i为0
}
}
return 0;
}
void qs(int l, int r)
{
if(l >= r) return ;
int i = l, j = r;
struct node t = b[l];
while(i < j)
{ // 开始就是这里出问题了，注释掉的部分是失败的。。。但是不清楚为什么，看得出来的小伙伴欢迎下方评论指正
while(i < j && b[j].data <= t.data /*||(b[j].data == t.data && b[j].x < t.x))*/)
{
j--;
}
b[i] = b[j];
while(i < j && b[i].data >= t.data /*||(b[i].data == t.data && b[i].x > t.x))*/)
{
i++;
}
b[j] = b[i];
}
b[i] = t;
qs(l, i - 1);
qs(i + 1, r);
}