Codeforces Round #524 (Div. 2)

好不容易考完电路，又可以回来刷题了，只是现在恐怕码力连新生都比不上了。

A	Petya and Origami

开头就是一大水题，意思就是制作一张贺卡需要几张什么颜色的纸，一共做n份，一叠纸有k张纸，问你需要几叠纸。

1 #include<iostream>
2 #include<cmath>
3 using namespace std;
4 int main()
5 {
6     int n,k;
7     scanf("%d%d",&n,&k);
8     printf("%d
",(int)ceil(1.0*n*8/k)+(int)ceil(1.0*n*5/k)+(int)ceil(1.0*n*2/k));
9 }

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B	Margarite and the best present

给你一个数组，-1，2，-3，4.。。。。。，q次询问，求l到r之间的数组和

直接算就可以了。相邻的两个差距为1嘛！

 1 #include<iostream>
 2 using namespace std;
 3 int main()
 4 {
 5      int T;
 6      scanf("%d",&T);
 7      while(T--){
 8         int l,r;
 9         scanf("%d%d",&l,&r);
10         int flag1,flag2;
11         if(!(l&1)){l--;flag1=l;}
12         else flag1=0;
13         if(r&1){r++;flag2=r;}
14         else flag2=0;
15         printf("%d
",(r-l+1)/2+flag1-flag2);
16      }
17 }

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C	Masha and two friends

过程稍稍有些麻烦，但是确实不难，就是那个求两矩形相交的公式是新学的，我实在是太弱了。

中间有一个过程，(a+b)%2==0换成（a+b)&1==0就会出问题，虽然我也不知道为什么

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<vector>
 4 #include<stack>
 5 #include<queue>
 6 #include<map>
 7 #include<set>
 8 #include<cstdio>
 9 #include<cstring>
10 #include<cmath>
11 #include<ctime>
12 #define fuck(x) cout<<#x<<" = "<<x<<endl;
13 #define ls (t<<1)
14 #define rs ((t<<1)+1)
15 using namespace std;
16 typedef long long ll;
17 typedef unsigned long long ull;
18 const int maxn = 100086;
19 const int inf = 2.1e9;
20 const ll Inf = 999999999999999999;
21 const int mod = 1000000007;
22 const double eps = 1e-6;
23 const double pi = acos(-1);
24 
25 ll cal(ll x,ll y)
26 {
27     if(x%2==0||y%2==0){return x*y/2;}
28     return x*y/2+1;
29 }
30 
31 int main()
32 {
33 
34     int T;
35     scanf("%d",&T);
36     ll a,b,c,d,e,f,g,h;
37     ll white,black;
38     ll n,m;
39     while(T--){
40         scanf("%lld%lld",&n,&m);
41         white = cal(n,m);
42         black = n*m-white;
43         scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
44         ll t = cal(c-a+1,d-b+1);
45         if((a+b)%2==0){t = (c-a+1)*(d-b+1)-t;}
46         white+=t;
47         black-=t;
48 
49         scanf("%lld%lld%lld%lld",&e,&f,&g,&h);
50         t = cal(g-e+1,h-f+1);
51         if((e+f)&1){t = (g-e+1)*(h-f+1)-t;}
52         white-=t;
53         black+=t;
54 
55         a = max(a,e);b=max(b,f);c=min(c,g);d=min(d,h);
56         t = cal(c-a+1,d-b+1);
57         if((a+b)%2==0){t = (c-a+1)*(d-b+1)-t;}
58         if(c<a||d<b){t=0;}
59         white-=t;
60         black+=t;
61         printf("%lld %lld
",white,black);
62     }
63 
64 
65     return 0;
66 }

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