URAL 1998 The old Padawan 二分

The old Padawan

题目连接：

http://acm.timus.ru/problem.aspx?space=1&num=1998

Description

Yoda: Use the Force. Yes. Now, the stone. Feel it. Concentrate!
Luke Skywalker is having exhausting practice at a God-forsaken planet Dagoba. One of his main difficulties is navigating cumbersome objects using the Power. Luke’s task is to hold several stones in the air simultaneously. It takes complete concentration and attentiveness but the fellow keeps getting distracted.
Luke chose a certain order of stones and he lifts them, one by one, strictly following the order. Each second Luke raises a stone in the air. However, if he gets distracted during this second, he cannot lift the stone. Moreover, he drops some stones he had picked before. The stones fall in the order that is reverse to the order they were raised. They fall until the total weight of the fallen stones exceeds k kilograms or there are no more stones to fall down.
The task is considered complete at the moment when Luke gets all of the stones in the air. Luke is good at divination and he can foresee all moments he will get distracted at. Now he wants to understand how much time he is going to need to complete the exercise and move on.

Input

The first line contains three integers: n is the total number of stones, m is the number of moments when Luke gets distracted and k (1 ≤ n, m ≤ 10 5, 1 ≤ k ≤ 10 9). Next n lines contain the stones’ weights wi (in kilograms) in the order Luke is going to raise them (1 ≤ wi ≤ 10 4). Next m lines contain moments ti, when Luke gets distracted by some events (1 ≤ ti ≤ 10 9, ti < ti+1).

Output

Print a single integer — the number of seconds young Skywalker needs to complete the exercise.

Sample Input

5 1 4
1
2
3
4
5
4

Sample Output

8

Hint

题意

这个人每秒钟会捡起来一个物品

这个人会在某些时间发呆，如果发呆的话，他会一直丢下物品，直到丢完或者丢下价值和超过k的物品

丢下去的顺序是捡起来顺序的倒叙

问你最少多少秒，可以把所有东西都捡起来

题解：

对于发呆这个东西，直接二分就好了，二分一直丢到哪儿

代码

#include<bits/stdc++.h>
using namespace std;
const long long inf = 1LL<<50;
const int maxn = 1e5+7;
long long sum[maxn],t[maxn],a[maxn],k;
int n,m;
int main()
{
    scanf("%d%d%lld",&n,&m,&k);
	for(int i=1;i<=n;i++)
		scanf("%lld",&a[i]);
	for(int i=n;i>=1;i--)
		sum[i]=sum[i+1]+a[i];
	for(int i=1;i<=m;i++)
		scanf("%lld",&t[i]);
	t[m+1] = inf;
	sum[0] = inf;
	int now = 0;
	long long ans = 0;
	int j = 1;
	while(1)
	{
		if(j==m+1)
		{
			ans = ans + n - now;
			break;
		}
		if(t[j]>ans+n-now)
		{
			ans = ans + n - now;
			break;
		}
		now = now + (t[j]-t[j-1]) - 1;
		int l = 0,r = now,Ans = 1;
		while(l<=r)
		{
			int mid = (l+r)/2;
			if(sum[mid]-sum[now+1]>k)Ans=mid,l=mid+1;
			else r=mid-1;
		}
		now = Ans - 1;
		now = max(now,0);
		ans = t[j];
		j++;
	}
	cout<<ans<<endl;
}