Ural 1998 The old Padawan（二分）

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1998. The old Padawan

Time limit: 0.5 second

Memory limit: 64 MB

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 ≤ 105,
1 ≤ k ≤ 109). Next n lines contain the stones’ weights wi (in kilograms)
in the order Luke is going to raise them (1 ≤ wi ≤ 104). Next m lines contain
moments ti, when Luke gets distracted by some events (1 ≤ ti ≤
109, ti < ti+1).

Output

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

Sample

input	output
5 1 4 1 2 3 4 5 4	8

Hint

In the first three seconds Luke raises stones that weight 1, 2 and 3 kilograms. On the fourth second he gets distracted and drops stones that weight 2 and 3 kilograms. During the next four seconds he
raises all the four stones off the ground and finishes the task.

题意：有一个人要举起m块石头，但是在举石头的过程中容易走神，在他走神的1s内，要向前掉至少Kkg的石头，求其至少多少步才能把石头都举起来。

二分判断现在举起的石头还有多少。然后一步一步来。



代码：

#include <cstdio>
#include <cstring>
#include<iostream>
#include<queue>
#include <algorithm>
#define N 100005
using namespace std;
int n,m,k;
int w
,t
,s
;
int binary(int x)
{
int l=0,r=n,mid,ss=0;
while(l<=r)
{
mid=(l+r)/2;
if(s[mid]<x)
{ l=mid+1;ss=mid;}
else
r=mid-1;
}
return ss;
}
int main()
{
while(~scanf("%d%d%d",&n,&m,&k))
{
memset(s,0,sizeof(s));
memset(w,0,sizeof(s));
memset(t,0,sizeof(s));

for(int i=1;i<=n;i++)
{
scanf("%d",&w[i]);
s[i]=s[i-1]+w[i];
}
for(int i=1;i<=m;i++)
scanf("%d",&t[i]);
t[++m]=2e9;

int sum=0,tmp=0,p=0;
for(int i=1;i<=m;i++)
{
if(n-p<t[i]-sum)
{
sum+=n-p;break;
}
else
{
tmp=t[i]-sum+p;
sum+=t[i]-sum;
p=binary(s[tmp-1]-k);
}
}
printf("%d\n",sum);
}
return 0;
}