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487-3279的解法实例
//解法1 #include<iostream> #include<string> using namespace std; int convert(char a) { int flag=-1; if(a==‘A‘||a==‘B‘||a==‘C‘||a==‘2‘) flag=2; if(a==‘D‘||a==‘E‘||a==‘F‘||a==‘3‘) flag=3; if(a==‘G‘||a==‘H‘||a==‘I‘||a==‘4‘) flag=4; if(a==‘J‘||a==‘K‘||a==‘L‘||a==‘5‘) flag=5; if(a==‘M‘||a==‘N‘||a==‘O‘||a==‘6‘) flag=6; if(a==‘P‘||a==‘R‘||a==‘S‘||a==‘7‘) flag=7; if(a==‘T‘||a==‘U‘||a==‘V‘||a==‘8‘) flag=8; if(a==‘W‘||a==‘X‘||a==‘Y‘||a==‘9‘) flag=9; if(a==‘1‘) flag=1; if(a==‘0‘) flag=0; return flag; } char convert1(int a) { char f; if(a==0) f=‘0‘; if(a==1) f=‘1‘; if(a==2) f=‘2‘; if(a==3) f=‘3‘; if(a==4) f=‘4‘; if(a==5) f=‘5‘; if(a==6) f=‘6‘; if(a==7) f=‘7‘; if(a==8) f=‘8‘; if(a==9) f=‘9‘; return f; } int main() { string s; int p,q; string temp; int temp1; string a[10000]; for(p=0;p<10000;p++) a[p]=""; int sum; cin>>sum; int count=0; for(int i=0;i<sum;i++) { cin>>s; int j; int count1=0; for(j=0;j<s.length();j++) { temp1=convert(s[j]); if(temp1!=-1) { if(count1==3) { a[count]=a[count]+‘-‘; count1++; } a[count]=a[count]+convert1(temp1); count1++; } } count++; } int w[10000]; string b[10000]; int c[10000]; for(int i=0;i<10000;i++) b[i]=""; for(int i=0;i<count;i++) w[i]=1; int u=0; for(int i=0;i<count;i++) { if(w[i]==1) { int count2=1; for(int j=i+1;j<count;j++) { if(a[i]==a[j]) { w[j]=0; count2++; } } if(count2>1) { b[u]=a[i]; c[u]=count2; u++; } } } int d[100000]; for(int z=0;z<u;z++) { d[z]=convert(b[z][0]); } int min; int y=0; for(int r=0;r<u;r++) { min=d[y]; for(int z=0;z<u;z++) { if(d[z]<min) { min=d[z]; y=z; } } cout<<b[y]<<" "<<c[y]<<endl; d[y]=100000; } return 0; } //解法2 #include <iostream> #include <map> #include <string> using namespace std; int main() { char szNumMap[] = "22233344455566670778889990"; map<string,int> mResultMap; string strInputTemp; string strResult; int nCaseCount = 0; cin >> nCaseCount; while( nCaseCount > 0 ) { cin >> strInputTemp; strResult.clear(); for( string::size_type i = 0; i < strInputTemp.size(); ++i ) { if ( strInputTemp[i] == ‘-‘ ) continue; if ( isdigit( strInputTemp[i] ) ) { strResult += strInputTemp[i]; } else { strResult += szNumMap[strInputTemp[i]-‘A‘]; } } strResult.insert( 3, 1, ‘-‘ ); mResultMap[strResult]++; --nCaseCount; } bool bRepeat = false; for ( map<string,int>::const_iterator it = mResultMap.begin(); it != mResultMap.end(); ++it ) { if ( it->second > 1 ) { cout << it->first << " " << it->second << endl; bRepeat = true; } } if ( !bRepeat ) { cout << "No duplicates." << endl; } return 0; } //解法3 typedef struct { int identy; char str[9]; int icount; }Phone; #include <stdio.h> #include <stdlib.h> #define MAXSIZE 100000 int cmp(const void *a, const void *b) { int *x = (int *)a; int *y = (int *)b; return (*x) - (*y); } int main() { int cnt = 0; //存放电话号码个数 int iter = 0; int Noduplicates = 0; //有无重复号码的标志,为0则表示无重复 int Sum = 0; int icount = 0; char ch; // 接收键盘输入字符 int *PHONE = (int *)malloc(sizeof(int) * (MAXSIZE + 10)); scanf("%d",&cnt); getchar(); //获取所有电话号码,转换为整型数据 for(iter = 0; iter < cnt; iter++) { Sum = 0; ch = getchar(); while(ch != ‘\n‘) { if(ch >= ‘0‘ && ch <= ‘9‘) { Sum = Sum * 10 + ch - ‘0‘ ; } else if(ch >= ‘A‘ && ch <= ‘P‘) { Sum = Sum * 10 + (ch - 65) / 3 + 2 ; } else if(ch >= ‘R‘ && ch <= ‘Y‘) { Sum = Sum * 10 + (ch - 66) / 3 + 2 ; } ch = getchar(); } PHONE[iter] = Sum; } //用快排对电话号码排序 qsort(PHONE, cnt, sizeof(int), cmp); //输出排好顺序的号码 for(iter = 0, icount = 1; iter < cnt - 1 ; iter++) { icount = 1; //统计重复号码个数 while(PHONE[iter] == PHONE[iter + 1]) { iter++; icount++; } if(icount > 1) { Noduplicates = 1; printf("%03d-%04d %d\n",PHONE[iter] / 10000 ,PHONE[iter] % 10000 ,icount); } } if(Noduplicates == 0) { printf("No duplicates.\n"); } getchar(); return 0; } //解法4 #include<stdio.h> #include<string.h> #include <stdlib.h> int cmp(const void *a, const void *b){ return(*(int *)a-*(int *)b); } int main(){ char str[100]; int num[26]={2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 0, 7, 7, 8, 8, 8, 9, 9, 9, 0}; int n, a[100005], l, sum, i, k, flag, t, s, j, p; scanf("%d", &n); getchar(); k = 0; while (k!=n){ gets(str); l = strlen(str); sum = 0; for (i=0; i<l; i++){ if (str[i]>= ‘0‘ && str[i] <= ‘9‘){ sum = sum*10 + str[i]-‘1‘+1; continue; } if (str[i]>=‘A‘ && str[i]<=‘Z‘){ if (str[i] == ‘Q‘ || str[i] == ‘Z‘) continue; sum = sum * 10 + num[str[i]-‘A‘]; } } a[k] = sum; k++; } a[k] = -1; qsort(a, k, sizeof(a[0]), cmp); t = a[0]; flag = 0; i = 1; s = 1; while (i!=k+1){ if (t == a[i]){ s++; flag = 1; }else { if (s>1){ p = 1000000; for(j=1; j<=3; j++){ printf("%d", t/p); t = t%p; p = p/10; } printf("-"); p = 1000; for (j=1; j<=4; j++){ printf("%d", t/p); t = t%p; p = p/10; } printf(" %d\n", s); } t = a[i]; s=1; } i++; } if (flag == 0){ printf("No duplicates.\n"); } return 0; }
487-3279的解法实例
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