#291 div.2

A.水题 数字翻转，将每一位大于等于5的数字t翻转成9-t，注意不要有前导0，且翻转后数字的位数不变（即9999->9000…刚开始以为应该翻转成0了= =）

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 using namespace std;
 int main()
 {
     char x[];
     cin >> x;
     int len = strlen(x);
     for(int i = ; i < len; i++)
         if(x[i] >= '') x[i] = '' - x[i] + '';
     if(x[] == '') x[] = '';
     cout << x << endl;
     return ;
 }
 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 using namespace std;
 int main()
 {
     char x[];
     cin >> x;
     int len = strlen(x);
     for(int i = ; i < len; i++)
         if(x[i] >= '') x[i] = '' - x[i] + '';
     if(x[] == '') x[] = '';
     cout << x << endl;
     return ;
 }

B.水题 斜率，每输入一个坐标(xi,yi)求出该点与标准点连线的斜率并记入数组k[]中，排序（sort），去重（unique）；

NOTE：斜率注意的基本问题—斜率为无穷大与0时的特判；

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int maxn = ;
 int main()
 {
     int n, cnt1 = , cnt2 = ;
     double x0, y0, x, y;
     double k[maxn];
     scanf("%d%lf%lf", &n, &x0, &y0);
     for(int i = ; i < n; i++)
     {
         scanf("%lf%lf", &x, &y);
         if(x-x0 == ) {cnt1++; k[i] = ; continue;}
         if(y-y0 == ) {cnt2++; k[i] = ; continue;}
         k[i] = (y-y0)/(x-x0);
     }
     sort(k, k+n);
     int ans = unique(k, k+n) - k;
     if(cnt1 && cnt2) ans++;
     cout << ans << endl;
     return ;
 }