• 统计字母个数


    1164: 统计字母个数

    Time Limit: 1 Sec  Memory Limit: 128 MB
    Submit: 890  Solved: 437
    [Submit][Status][BBS]

    Description

    给定一段文章,请输出每个字母出现的次数

    Input

    只有一组输入数据,该数据大小<10KB。在文章中除最后一个字符外,只有小写字母、空格和换行符,没有另外的标点、数字和大写字母等。该文章以’#’结尾。

    Output

    输出格式为“C A”,C为’a’..’z’中的字母,A为出现次数,C和A之间空一格

    Sample Input

    here is the input
    this is the article#

    Sample Output

    a 1
    b 0
    c 1
    d 0
    e 5
    f 0
    g 0
    h 4
    i 5
    j 0
    k 0
    l 1
    m 0
    n 1
    o 0
    p 1
    q 0
    r 2
    s 3
    t 5
    u 1
    v 0
    w 0
    x 0
    y 0
    z 0

    HINT

    Source

    吉首大学软件学院

    #include <stdio.h>
    #include <string.h>
    char a[13000];
    int  main()
    {
        int A=0,B=0,C=0,D=0,E=0,F=0,G=0,H=0,I=0,J=0,K=0,L=0,M=0,
            N=0,O=0,P=0,Q=0,R=0,S=0,T=0,U=0,V=0,W=0,X=0,Y=0,Z=0;
        while(1)
        {
            gets(a);//遇到回车进行一次计数,但还要进入循环再次输入
            int flag = 0;
            int len=strlen(a);
            for(int i=0; i<len; i++)
            {
                if(a[i]=='#')
                {
                    flag=1;
                    break;
                }
                if(a[i]=='a') A++;
                if(a[i]=='b') B++;
                if(a[i]=='c') C++;
                if(a[i]=='d') D++;
                if(a[i]=='e') E++;
                if(a[i]=='f') F++;
                if(a[i]=='g') G++;
                if(a[i]=='h') H++;
                if(a[i]=='i') I++;
                if(a[i]=='j') J++;
                if(a[i]=='k') K++;
                if(a[i]=='l') L++;
                if(a[i]=='m') M++;
                if(a[i]=='n') N++;
                if(a[i]=='o') O++;
                if(a[i]=='p') P++;
                if(a[i]=='q') Q++;
                if(a[i]=='r') R++;
                if(a[i]=='s') S++;
                if(a[i]=='t') T++;
                if(a[i]=='u') U++;
                if(a[i]=='v') V++;
                if(a[i]=='w') W++;
                if(a[i]=='x') X++;
                if(a[i]=='y') Y++;
                if(a[i]=='z') Z++;
            }
            if(flag==1)//连续跳出两次才跳得出
                break;
        }
        printf("a %d
    ",A);
        printf("b %d
    ",B);
        printf("c %d
    ",C);
        printf("d %d
    ",D);
        printf("e %d
    ",E);
        printf("f %d
    ",F);
        printf("g %d
    ",G);
        printf("h %d
    ",H);
        printf("i %d
    ",I);
        printf("j %d
    ",J);
        printf("k %d
    ",K);
        printf("l %d
    ",L);
        printf("m %d
    ",M);
        printf("n %d
    ",N);
        printf("o %d
    ",O);
        printf("p %d
    ",P);
        printf("q %d
    ",Q);
        printf("r %d
    ",R);
        printf("s %d
    ",S);
        printf("t %d
    ",T);
        printf("u %d
    ",U);
        printf("v %d
    ",V);
        printf("w %d
    ",W);
        printf("x %d
    ",X);
        printf("y %d
    ",Y);
        printf("z %d
    ",Z);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/tianmin123/p/4859253.html
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