说说:这道题初看挺简单的,题意无非就是将一串字符转化成一个用‘-’隔开的电话号码,然后把出现超过一次的号码按照字典升序输出即可。但是这样样做是会超时的....其实把电话号码中间的‘-’去掉,电话号码其实就是一个整数,有了这个想法那就简单啦。只要设立一个超大的数组包含所有可能的电话号码,然后数组的值是该号码出现的次数,统计完后遍历一遍输出即可。但是第三种相当于把前两种方法法结合起来了。把号码当成一个整数,但是号码存储在一个数组中,号码出现的次数存储在另一个数组中。这样在插入新号码的时候就排序,然后就超时了。总的来说,不能排序,因为数据量比较大,最后肯定会超时的。
题目:
487--3279
Businesses like to have memorable telephone numbers. One way to make a telephone number memorable is to have it spell a memorable word or phrase. For example,
商务人士都喜欢容易记住的电话号码。让号码听起来像一个易记的单词或短语是一种很好的方法。
you can call the University of Waterloo by dialing the memorable TUT-GLOP. Sometimes only part of the number is used to spell a word. When you get back to your hotel
比如你打Waterloo大学的电话就拨TUT-GLOP。有时只有一部分的数字用于组成一个单词。当你在今天晚上回到你的酒店的时候
tonight you can order a pizza from Gino‘s by dialing 310-GINO. Another way to make a telephone number memorable is to group the digits in a memorable way. You
你可以拨打310-GINO从GION预订披萨。另一种让电话号码易记的方法是用一种易记的方式组织数字。
could order your pizza from Pizza Hut by calling their ``three tens‘‘ number 3-10-10-10.
你可以从PIZZA HUT预订披萨通过拨打三个十------ 3-10-10-10
The standard form of a telephone number is seven decimal digits with a hyphen between the third and fourth digits (e.g. 888-1200). The keypad of a phone supplies the
标准的电话号码形式是七个十进制数,并且在第三个和第四个数之间有一个连字符(例如:888-1200).一个手机的键盘提供了一种从字母到数字的映射
mapping of letters to numbers, as follows:
如下所示:
A, B, and C map to 2
D, E, and F map to 3
G, H, and I map to 4
J, K, and L map to 5
M, N, and O map to 6
P, R, and S map to 7
T, U, and V map to 8
W, X, and Y map to 9
There is no mapping for Q or Z. Hyphens are not dialed, and can be added and removed as necessary.The standard form of TUT-GLOP is 888-4567, the standard form of
其中并没有到Q或Z的映射。连字符在拨号时不必使用,在必要的时候可用可不用。TUT-GLOP的标准形式是888-4567,310-GINO的标准形式是310-4466,3-10-10-10的标准形
310-GINO is 310-4466, and the standard form of 3-10-10-10 is 310-1010.
式是310-1010.
Two telephone numbers are equivalent if they have the same standard form. (They dial the same number.)
如果两个电话号码有同样的标准形式,那么它们就是完全等价的。
Your company is compiling a directory of telephone numbers from local businesses. As part of the quality control process you want to check that no two (or more)
你的公司现在正在编写一部关于本地商业信息的电话号码本。作为质量控制,你现在要确保在电话本中没有两个(或更多)的电话是一样的。
businesses in the directory have the same telephone number.
INPUT
The first line of the input contains the number of data sets in the input. A blank line follows. The first line of each data set specifies the number of telephone numbers in the
第一行的输入是输入数据集的个数。接下来会有一个空白行。每个输入数据集的第一行是一个正数,指明了电话本中电话号码的数目(最大为100,000)。
directory (up to 100,000) as a positive integer alone on the line. The remaining lines list the telephone numbers in the directory, with each number alone on a line. Each
接下来每一行是电话本中的一个号码。
telephone number consists of a string composed of decimal digits, uppercase letters (excluding Q and Z) and hyphens. Exactly seven of the characters in the string will be
每个电话号码由十进位数字,大写字母(除了Q和Z)以及连字符组成且数字和字母的总数刚好是七个。
digits or letters. There‘s a blank line between data sets.
每个数据集直接有一个空行。
OUTPUT
Generate a line of output for each telephone number that appears more than once in any form. The line should give the telephone number in standard form, followed by a
首先将那些以任何形式出现过一次以上的电话号码以标准形式输出,并且接下来输出一个空格,接着输出电话号码在电话本中出现的次数。
space, followed by the number of times the telephone number appears in the directory. Arrange the output lines by telephone number in ascending lexicographical order. If
而且输出的电话号码必须按字典升序输出。
there are no duplicates in the input print the line:
如果没有重复的电话号码则输出:
No duplicates.
Print a blank line between data sets.
在数据集的输出直接输出一行空行。
SAMPLE INPUT
1 12 4873279 ITS-EASY 888-4567 3-10-10-10 888-GLOP TUT-GLOP 967-11-11 310-GINO F101010 888-1200 -4-8-7-3-2-7-9- 487-3279
SAMPLE OUTPUT
310-1010 2 487-3279 4 888-4567 3
源代码:
解法2:(AC过的,但还是要600多毫秒= =)
#include <stdio.h> #include <string.h> #include <ctype.h> #define MAXN 10000000+5 int tel[MAXN];//数组下标是电话号码,数组值是下标对应的号码出现的次数 int N,M,count; char new_tel[50]; int to_standard(char*);//电话转化为数字 char AtoD(char);//字母转化为数字 int main(){ int i,pos,val; // freopen("data","r",stdin); scanf("%d",&M); while(M--){ scanf("%d",&N); count=0; memset(tel,0,sizeof(tel)); for(i=0;i<N;i++){ scanf("%s",new_tel); val=to_standard(new_tel); if(tel[val]==0)//号码未出现过 count++; tel[val]++; } if(count==N)//号码每个都不同 printf("No duplicates.\n"); else for(i=0;i<MAXN;i++) if(tel[i]>1) printf("%03d-%04d %d\n",i/10000,i%10000,tel[i]); if(M) putchar('\n'); } return 0; } int to_standard(char *new_tel){//字符号码转化为整数值 int i=0; int val=0; while(new_tel[i]){ if(isdigit(new_tel[i])) val=val*10+new_tel[i++]-'0'; else if(new_tel[i]=='-'){ i++; continue; } else val=val*10+AtoD(new_tel[i++])-'0'; } return val; } char AtoD(char c){//字母转化为数字 if(c>'Q') c--; return '2'+(c-'A')/3; }
为了保证文章的长度,另外两种方法就不贴代码啦。嘿嘿~
487--3279 UVA 755 其实有三种解法,布布扣,bubuko.com