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。

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

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！

#include<iostream>
using namespace std;
int main()
{
     int T;
     scanf("%d",&T);
     while(T--){
        int l,r;
        scanf("%d%d",&l,&r);
        int flag1,flag2;
        if(!(l&1)){l--;flag1=l;}
        else flag1=0;
        if(r&1){r++;flag2=r;}
        else flag2=0;
        printf("%d
",(r-l+1)/2+flag1-flag2);
     }
}

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

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100086;
const int inf = 2.1e9;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);

ll cal(ll x,ll y)
{
    if(x%2==0||y%2==0){return x*y/2;}
    return x*y/2+1;
}

int main()
{

    int T;
    scanf("%d",&T);
    ll a,b,c,d,e,f,g,h;
    ll white,black;
    ll n,m;
    while(T--){
        scanf("%lld%lld",&n,&m);
        white = cal(n,m);
        black = n*m-white;
        scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
        ll t = cal(c-a+1,d-b+1);
        if((a+b)%2==0){t = (c-a+1)*(d-b+1)-t;}
        white+=t;
        black-=t;

        scanf("%lld%lld%lld%lld",&e,&f,&g,&h);
        t = cal(g-e+1,h-f+1);
        if((e+f)&1){t = (g-e+1)*(h-f+1)-t;}
        white-=t;
        black+=t;

        a = max(a,e);b=max(b,f);c=min(c,g);d=min(d,h);
        t = cal(c-a+1,d-b+1);
        if((a+b)%2==0){t = (c-a+1)*(d-b+1)-t;}
        if(c<a||d<b){t=0;}
        white-=t;
        black+=t;
        printf("%lld %lld
",white,black);
    }


    return 0;
}