2013年华为机考参考代码8-29

/************************************************************************//* 2013-8-29华为机考 第三题，这个由于是最后一题，会比较难点；前面两题比较简单的给定两个字符串，设定每个字符的权值(1-26),字母大小写一样，求出两字符串的最大差值；思路：先去相同的字母，然后分别两个字符串map<char,int,greater<>>,长串是sum1，求出最大sum1和最小sum2，max=sum1-sum2就是答案了。author 2013-8-29-15:00*//************************************************************************/#include <iostream>#include <map>#include <set>#include <iterator>using namespace std;char s1[300];char s2[300];void Max_large_diff(char *s1,char *s2,int len2){int i,j,k;char temp;int sum1,sum2;int sum=0;int finish=0;map<char,int,greater<int>>map_char1;map<char,int,greater<int>>map_char2;map<char,int,greater<int>>::iterator it;sum1=0;sum2=0;for(i=0;s1[i]!=0;i++){temp =s1[i];for(j=0;s2[j]!=0 && finish!=len2;j++){if(tolower(temp) == tolower(s2[j])){s1[i]=-1;s2[j]=-1;finish++;break;}}}for (i =0;s1[i]!=0;i++)if(s1[i]!=-1)map_char1[s1[i]]++;for(i =0;s2[i]!=0;i++)if(s2[i]!=-1)map_char2[s2[i]]++;it = map_char1.begin();k=26;for(;it!=map_char1.end();it++){sum1 = sum1 + (*it).second *k ;k--;}k = 1;it = map_char2.begin();for(;it!=map_char2.end();it++){sum2 = sum2 + (*it).second *k;k++;}sum = sum1 - sum2;printf("%d\n",sum);}int main(){while(scanf("%s",s1)!=EOF){scanf("%s",s2);int n1= strlen(s1);int n2 = strlen(s2);if(n1 >= n2){Max_large_diff(s1,s2,n2);}else<pre name="code" class="cpp">/************************************************************************//* 2013-8-29华为机考 第二题，一堆人按顺序报数，报到N，就出列，继续，问最后一个出列的是多少？直接模拟游戏，得出结果！author 2013-8-29-15:00*//************************************************************************/#include <iostream>using namespace std;int main(){int s[100];int N,i,j;int number;int last_N;int count_N;while(scanf("%d",&N)!=EOF){memset(s,0,sizeof(s));count_N =0;j = 0;if(N==1){printf("1\n");continue;}while(1){count_N++;j++;if(j==(N+1))j=1;while(s[j]!=0){j++;if(j==(N+1))j=1;}if(count_N %3==0){s[j] = 1;number = 0;for(int i=1;i<=N;i++){if(s[i]==0){number++;last_N = i;}}if(number ==1){printf("%d\n",last_N);break;}count_N=0;}}}return 0;}

{Max_large_diff(s2,s1,n1);}}return 0;}

/************************************************************************//* 2013-8-29华为机考 第一题，字符串的大小写转化，相当简单！author 2013-8-29-15:00*//************************************************************************/#include <iostream>using namespace std;int main(){char s[1000];int n;while(scanf("%s",s)!=EOF){for(n =0;s[n]!=0;n++){if(s[n]>='a' && s[n]<='z')s[n] = s[n] - ('a'-'A');else if(s[n]>='A' && s[n]<='Z')s[n] = s[n] + ('a'-'A');}printf("%s\n",s);}return 0;}