华为机试网上练习二

1、第一题（60分）:按要求分解字符串，输入两个数M , N ; M代表输入的M串字符串，N代表输出的每串字符串的位数，不够补0。例如：输入2,8，a I »abc“ 123456789 ” 则输出为 “ abc00000 ” , “ 12345678“ , ”90000000 ”cpp view plaincopy1.转载请标明出处，原文地址：htt P： /2.#include<iostream>3.#include<cstdio>4.using names pace std;5.6.void solve( char *str,int n , int7.8.i

2、nt i , j , k , quotient , remainder;9.quotient = len / n;10.remainder = len - n * quotient;11.12.for (i = 0 ; i < len ; i += n)13.14.if (len - i < n)15.16.k = n - len + i;17.for (j =i ; j < len ; +j)18.p rintf("%c" , strj)19.for (j =0 ; j < k ; +j)20.pu tchar('0');21.22

3、.else23.24.for (j = i;j < i + n ; +j)25.p rintf("%c" , strj);26.27.pu tchar('');28.29.printf("n");30.31.32.int main( void )J33.len)/原字符串被分解的个数/剩余的字符串的个数34.int i,m , n , len;35.charstr1000;36.37.while(scanf( "%d %d" , &m38.39.for (i = 0 ; i < m ; +i)40.

4、41.scanf("%s" , str);42.len=strlen(str);43.solve(str , n , len);44.45.46.return 0;47.,&n) != EOF)第一题：拼音转数字输入是一个只包含拼音的字符串,请输出对应的数字序列。转换关系如下:描述:拼音yi ersan si wu liu qi ba jiu阿拉伯数字456789cpp view plaincopy输入字符只包含小写字母，所有字符都可以正好匹配运行时间限制:无限制内存限制:无限制输入:一行字符串，长度小于1000输出:一行字符（数字）串样例输入:yiersa nsi

5、样例输出:12341.转载请标明出处，原文地址:htt P： /2.#include<iostream>3.#include<cstdio>4.using names pace std;5.6.voidsolve( char *str ,intlen)7.8.inti;9.10.for(i = 0 ; i < len ;)11.12.switch (stri)13.14.case 'y':15.p utchar('1');16.i += 2;17.break ;18.case 'e19.p utchar('2'

6、;);20.i += 2;21.break ;22.case 's'23.if (stri +1='a')24.25.pu tchar('3');26.i += 3;27.28.else29.30.pu tchar('4');31.i += 2;32.33.break ;34.case 'w'35.p utchar('5');36.i += 2;37.break ;38.case 'l'39.p utchar('6');40.i += 3;41.break ;42.c

7、ase 'q':43.p utchar('7');44.i += 2;45.break46.case 'b':47.p utchar(48.i += 2;49.break50.case j'51.p utchar(52.i += 3;53.break54.55.56.printf("n");57. 58.59. intmain( void )60. 61.int len;62.char str1OOO;63.64.while (scanf('65.66.len = strlen(str);67.solve(st

8、r , len);68. 69.return 0;70. "%s" , str) != EOF)8 );'9')；第二题：去除重复字符并排序运行时间限制:无限制内容限制:无限制输入:字符串输出:去除重复字符并排序的字符串样例输入:aabcdefff样例输出:abcdef1.转载请标明出处，原文地址：htt P： /2.#includeviostream>3.#includevcstdio>4.#includevmemory>5.using names pace std;6.7.voidsolve( char *str ,intlen)8.9

9、.int i , hash256;10.memset(hash , 0 ,sizeof(hash);11.12.for (i = 0 ; i v len ; +i)13.14.if (0 = hashstri)15.hashstri = 1;16.17.for (i = 0 ; i v 256 ; +i)18.19.if (0 != hashi)20.p utchar(i);21.22.printf("n");23.24.25.intmain( void )26.27.int len;28.char str1000;29.30.while (scanf( "%s&

10、quot; , str) != EOF)31.32.len = strlen(str);33.solve(str , len);34.35.return 0;J36.第三题：等式变换输入一个正整数X,在下面的等式左边的数字之间添加 +号或者-号，使得等式成立。比如: 12-34+5-67+89 = 5 1+23+4-5+6-7-8-9 = 5 请编写程序，统计满足输入整数的所有整数个数。18.22.输入:正整数，等式右边的数字输出:使该等式成立的个数样例输入:样例输出:21cpp view plaincopy1.转载请标明出处，原文地址:htt P： /2.#include<iostre

11、am>3.#include<cstdio>4.using names pace std;5.6.7.int op s21;const char sym3 = '+'8.int result , num;9.10.void dfs( int layer, int currentResult,int last Op, int lastSum)12.11.13.lastSum *= (layer > 9) ? 100 : 10;lastSum += layer;14.if (layer = 9)15.16.17.currentResult += (last O

12、p) ? (-1 * lastSum) : lastSum;if (currentResult = result)19.+num;20.printf("1");21.for(int i = 2 ; i <= 9 ; +i)23.if (symopsi-1 !=24.p rintf("%c " , symopsi-1);25.p rintf("%d", i);26.27.printf("=%dn", result);28.29.return30.31.32.op slayer = 2;dfs(layer + 1 , currentResult , last Op , lastSum);/Continue33.currentResult += (lastO p)? (-1 * lastSum) : lastSum;34.op slayer = 0;35.dfs(layer + 1 , currentResult , 0,0);/P lus36.op slayer = 1;37.dfs(layer