计蒜客 A1011 爱奇艺的自制节目(贪心)

题目链接：https://vjudge.net/problem/%E8%AE%A1%E8%92%9C%E5%AE%A2-A1011

题目大意：w和x节目的拍摄场地是固定的，问你怎么分配x和y节目的时间从而使得他们在两个场地拍摄时间的最大值最小。

　　如果y和z同时处理情况会变得很复杂，所以我们可以暴力枚举出y或z分配在两边的数量，然后再用贪心考虑剩下的那一个如何分配最好

#include<set>
#include<map>
#include<list>
#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define endl '\n'
#define rtl rt<<1
#define rtr rt<<1|1
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
#define maxx(a, b) (a > b ? a : b)
#define minn(a, b) (a < b ? a : b)
#define zero(a) memset(a, 0, sizeof(a))
#define INF(a) memset(a, 0x3f, sizeof(a))
#define IOS ios::sync_with_stdio(false)
#define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n")
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<ll, ll> P2;
const double pi = acos(-1.0);
const double eps = 1e-7;
const ll MOD =  1000000007LL;
const int INF = 0x3f3f3f3f;
const int _NAN = -0x3f3f3f3f;
const double EULC = 0.5772156649015328;
const int NIL = -1;
template<typename T> void read(T &x){
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
const int maxn = 1e3+10;
int arr[maxn];
int main(void) {
    IOS; int t;
    cin >> t;
    while(t--) {
        ll ew, ex, ey, ez, w, x, y, z;
        cin >> ew >> ex >> ey >> ez >> w >> x >> y >> z;
        ew *= w, ex *= x;
        if (y<z) {
            swap(y, z);
            swap(ey, ez);
        }
        ll ans = LLONG_MAX;
        for (int i = 0; i<=ey; ++i) {
            ll ta = ew+i*y, tb = ex+(ey-i)*y;
            if (ta<tb) swap(ta, tb);
            ll sum, tmp = (ta-tb)/z;
            if (tmp >= ez) sum = ta;
            else {
                tb += tmp*z;
                ll tz = ez-tmp;
                sum = tz&1 ? tb+(tz+1)/2*z : ta+tz/2*z;
            }
            ans = min(ans, sum);
        }
        cout << ans << endl;
    }
    return 0;
}