CF round #292 解题报告

A题：

给出一个坐标(a,b),和走的步数，问有没有可能从（0,0）走这么多步刚好到达(a,b).

水题，结果还是wa了一次，没有考虑到a<0或者b<0的情况。

 1 #include<cstdio>
 2 int main()
 3 {
 4     int a,b,s;
 5     while(scanf("%d%d%d",&a,&b,&s)!=EOF){
 6         if(a<0)
 7             a=-a;
 8         if(b<0)
 9             b=-b;
10         if(a+b>s)
11             printf("NO
");
12         else{
13             s=a+b-s;
14             if(s%2)
15                 printf("No
");
16             else
17                 printf("Yes
");
18         }
19     }
20     return 0;
21 }

View Code

B题:

主人公有n个男性朋友，为0到n-1，m个女性朋友，为0到m-1，有些朋友开心，有些不开心。

其中b个男性朋友是开心的，n-b个不开心，

g个女性朋友是开心的，m-g个不开心。

给出n，m，

b，和b个坐标

g，和g个坐标

操作：

　　In i-th day, Drazil invites $\text{[math]}$ -th boy and $\text{[math]}$ -th girl to have dinner together (as Drazil is programmer, i starts from 0). If one of those two people is happy, the other one will also become happy. Otherwise, those two people remain in their states. Once a person becomes happy (or if he/she was happy originally), he stays happy forever.

我是直接走一遍，最后再扫描一遍，检查是不是所有人都开心了。

 1 #include<cstdio>
 2 #include<cstring>
 3 int boy[110];
 4 int girl[110];
 5 int main()
 6 {
 7     int n,m;
 8     while(scanf("%d%d",&n,&m)!=EOF){
 9         memset(boy,0,sizeof(boy));
10         memset(girl,0,sizeof(girl));
11         int b,bb;
12         scanf("%d",&b);
13         for(int i=0;i<b;i++){
14             scanf("%d",&bb);
15             boy[bb]=1;
16         }
17         int g,gg;
18         scanf("%d",&g);
19         for(int i=0;i<g;i++){
20             scanf("%d",&gg);
21             girl[gg]=1;
22         }
23         int sum=n*m*n*m;
24 
25         for(int i=0;i<sum;i++){
26             if(boy[i%n]||girl[i%m])
27                 boy[i%n]=girl[i%m]=1;
28         }
29         bool flag=true;
30         for(int i=0;i<n;i++)
31         if(!boy[i]){
32             flag=false;
33             break;
34         }
35         if(flag){
36             for(int i=0;i<m;i++)
37             if(!girl[i]){
38                 flag=false;
39                 break;
40             }
41         }
42         if(flag)
43             printf("Yes
");
44         else
45             printf("No
");
46     }
47     return 0;
48 }

View Code