A - Valera and Fruits

最新推荐文章于 2018-06-03 11:17:00 发布

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本题描述了一个关于收集水果的问题，玩家Valera要在果实成熟的日子收集尽可能多的水果，但每天只能收集限定数量的水果。文章提供了AC代码实现，通过记录每一天成熟的水果数量，并在连续两天内尽可能多地收集水果。

A - Valera and Fruits

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

Valera loves his garden, where n fruit trees grow.

This year he will enjoy a great harvest! On the i-th tree b_i fruit grow, they will ripen on a day number a_i. Unfortunately, the fruit on the tree get withered, so they can only be collected on day a_i and day a_i + 1 (all fruits that are not collected in these two days, become unfit to eat).

Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more thanv fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?

Input

The first line contains two space-separated integers n and v(1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.

Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers a_i and b_i(1 ≤ a_i, b_i ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.

Output

Print a single integer — the maximum number of fruit that Valera can collect.

Sample Input

Input

2 3
1 5
2 3

Output

Input

Output

Hint

In the first sample, in order to obtain the optimal answer, you should act as follows.

On the first day collect 3 fruits from the 1-st tree.
On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
On the third day collect the remaining fruits from the 2-nd tree.

In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.

小结：

今天有给在我后面的同学放水了呢...水题+1...

这道题目...好像没什么好说的...

以下是AC代码：

#include<stdio.h>
#include<string.h>
int chase[3100],value;
void init()
{
for(int i=0;i<=3000;i++)
chase[i]=0;
}
int main()
{
int nitem,temp1,temp2,maxd,ans;
while(~scanf("%d%d",&nitem,&value))
{
init();
maxd=0;
ans=0;
for(int i=1;i<=nitem;i++)
{
scanf("%d%d",&temp1,&temp2);
chase[temp1]+=temp2;
maxd=temp1>maxd?temp1:maxd;
}
for(int i=1;i<=maxd+1;i++)
{
int temp=value;
for(int j=i-1;j<=i;j++)
{
if(chase[j]<=temp)
{
temp-=chase[j];
ans+=chase[j];
chase[j]=0;
}
else
{
chase[j]-=temp;
ans+=temp;
temp=0;
}
}
}
printf("%d\n",ans);
}
return 0;
}