牛客练习赛6

A题：

方法一：一元二次方程求解，但是会有精度误差，+1个点特判一下。

#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ll;

ll t;
ll x,y,z;

int main()
{
    //freopen("in.txt","r",stdin);
    ll n;
    scanf("%lld%lld",&n,&t);

    ll ans = 0;
    for(ll i = 1; i <= n; i++) {
        scanf("%lld%lld%lld",&x,&y,&z);
        ll cnt;
        if(y==0&&z==0) continue;
        if(z==0) cnt = t/(x+y);
        else if(z!=0) {
            ll sq = ((x+y)-z/2)*((x+y)-z/2) + 2*z*t;
            sq = sqrt(sq) + z/2 - (x+y);
            cnt = (sq/z);
        }

        if((cnt+1)*(x+y)+(cnt+1)*cnt/2*z<=t)
            cnt++;

        ll tmp = cnt*(x+y) + cnt*(cnt-1)/2*z;
        if(t-tmp>=x) {
            ans += (t-(cnt+1)*x);
        }
        else {
            ans +=(t-cnt*x-(t-tmp));
        }

    }
    cout<<ans<<endl;

    return 0;
}

方法二：二分

二分是很坑的，二分上界是2e9，中间结果爆数据类型。大佬的解法是处理一下mid，2e9 >= (mid-1)*mid/2*z，移项一下。

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

ll n,t,ans,res;
ll x,y,z;

int main()
{
    scanf("%lld%lld",&n,&t);
    for(ll i = 1ll; i <= n; i++) {
        scanf("%lld%lld%lld",&x,&y,&z);

        ll l = 1,r= 2000000000;
        ans = 0;
        while(l<=r) {
            ll mid = (l+r)>>1;
            if(2000000000/mid>=(mid-1)*z/2&&(1ll*(x+y)*mid+1ll*(mid-1)*mid/2*z)<=t)
            {
                l = mid+1;
                ans = mid;
            }
            else r = mid - 1;
        }

        ll tmp = (1ll*(x+y)*ans+1ll*(ans-1)*ans/2*z);
        ll tt = 0;
        tt += tmp - 1ll*ans*x;
        if(x<t-tmp) tt+=t-tmp-x;
        res+=tt;

    }

    printf("%lld\n",res);

    return 0;
}

D题：

当时脑子短路，不知道为什么要二分t，然后b全部都减t，处理a变与不变。当然感觉也是可以的，但是硬是90%。无语了~~~

看了题解，简直吐血~

贪心，先把a给变了，然后对应的max(0,b[i]-a[i])；

看来以后得大胆的贪心~~~

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MAXN = 200005;
ll a[MAXN],b[MAXN];

int main()
{
    int n,x;
    ll y;
    cin>>n>>x>>y;

    for(int i = 0; i < n; i++) scanf("%lld",&a[i]);
    for(int i = 0; i < n; i++) scanf("%lld",&b[i]);

    sort(a,a+n);
    sort(b,b+n);

    for(int i = 0; i < x; i++) if(a[i]<y) a[i] = y;

    sort(a,a+n);

    ll ans = 0;
    for(int i = 0; i < n; i++) {
        ans = max(ans,b[i]-a[i]);
    }

    printf("%lld\n",ans);

    return 0;
}