Project Euler 17

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今天木有好好复习……还有20天考试……心里还是挺慌的……

好吧~这道题蛋碎，考英语~

嗯……本来我在笨笨的一点一点写……直到我膝盖中了一箭，觉得这样写下去实在是太二逼了……算了~上网简单看一下。

还是jason大牛的……

import time start = time.time() S = [0,3,3,5,4,4,3,5,5,4,3,6,6,8,8,7,7,9,8,8]D = [0,3,6,6,5,5,5,7,6,6]H = 7T = 8 total = 0for i in range(1,1000):    c = i % 10 # singles digit    b = ((i % 100) - c) / 10 # tens digit    a = ((i % 1000) - (b * 10) - c) / 100 # hundreds digit     if a != 0:        total += S[a] + H # "S[a] hundred        if b != 0 or c != 0: total += 3 # "and"    if b == 0 or b == 1: total += S[b * 10 + c]    else: total += D[b] + S[c] total += S[1] + Telapsed = time.time() - start print "%s found in %s seconds" % (total,elapsed)

写的简洁明了啊~

哎……自愧不如……

我的上来了……

#include<stdio.h>//one to nineteenint S[20] = {0,3,3,5,4,4,3,5,5,4,3,6,6,8,8,7,7,9,8,8};//twenty thirdty ....int D[10] = {0,0,6,6,5,5,5,7,6,6};//hundredint H = 7;//thousandint T = 8;//andint N = 3;int main(){int i, a, b, c;int sum = 0;for (i=1; i<1000; i++){c = i%10;b = ((i%100) - c)/10;a = i/100;// greater than 100if (a != 0){sum +=S[a] + H;if (b != 0 || c != 0){sum += N;}}// [1 to 19]if (b == 0 || b == 1){sum += S[b*10 + c];} else{  //[20 to 99]sum += D[b] + S[c];}}//add one thousandsum += S[1] + T;printf("%d", sum);}

带了些许的注释~咩~

突然想到……其实我以前文章都可以加上pe的标签啊……啊……还可以把题目复制上，找到翻译的……复制上……提高人气啊……

算了……

最后赞美一个~sublime真的太好用了！！！不过用着别人的key……真是不爽……