[交流] Euler 工程 第廿二题: 姓的总分已有5人参与

Originally posted by huycwork at 2011-06-06 13:45:03:
维持最奇怪语言的地位还真是有压力啊，俺再来三行版：

print sum([(i+1)*sum([ord(c)-64 for c in s]) for i,s in enumerate(sorted(open('names.txt').read()[1:-1].split('","')))])

%% What is the total of all the name scores in the file?
% For example, when the list is sorted into alphabetical order,
% COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list.
% So, COLIN would obtain a score of 938 × 53 = 49714.
function result = euler22()
tic;
fid = fopen('D:\euler\names.txt');
names = fgets(fid); % 读取全部内容到字符串
fclose(fid);

names = strrep(names,'"',''); % 删除"
namelist = regexp(names,',','split'); % 用逗号分隔
sname = sort(namelist); % 排序

result = 0;
for i=1:length(sname) % 循环
    curname = lower(sname{i}); % 全小写
    curname = curname-repmat('a',1,length(curname)); % 与全a作差
    cursum = sum(curname+1); % 求和
    result = result+cursum*i; % 乘顺序计算score,累加
end
toc;
end

% Elapsed time is 0.307340 seconds.
% ans =
%    871198282

#!/usr/bin/perl
open F, "<", "names.txt";
$f = ;
@ns = sort eval ($f);
foreach(@ns){
    local (*v) = \$_;
    $v += ord($_) - ord('A') + 1 foreach(/(.)/g);
}
unshift @ns, 0;
$s += $_ * $ns[$_] foreach(1..@ns);
print $s;

#include
#include
#include


#define SIZE 50000

inline int cmp(const void *p1,const void *p2)
{
        return strcmp((char *)p1,(char *)p2);
}

inline int count(const char *p){
        int i=0,res=0;
        while(p[i]!='\0')res+=p[i++]-64;
        return res;
}

long euler22(){
        int i=0,length;
        FILE *fp=fopen("names.txt", "r");
        if(fp == 0) return -1;

        char str[SIZE];
        if (NULL == fgets(str, SIZE, fp)) {
                return -1;
        }
        fclose(fp);

        char *delims="\",";
        char *p=strtok(str,delims);
        char names[6000][15];
        while(p!=NULL){
                strcpy(names[i++],p);
                p=strtok(NULL,delims);
        }
        length=i;
  
        qsort(names,length,sizeof(names[0]),cmp);
  
        long  sum=0;
        for(i=0;i                 sum+=(i+1)*count(names[i]);
      
        return sum;
}



int main(void){

        printf("%ld\n",euler22());
  
        return 0;
}