Edward's Cola Plan

 Edward's Cola Plan

Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

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Description

Edward is a tall, handsome and rich (a.k.a GaoShuaiFu, GSF) student who studys in Zhejiang University. He also takes part in many competition of ACM-ICPC, and he is so smart that he almost wins the championship of World Finals recently! Unfortunately, he doesn't spent time in studying, so when the result of the final examination is published, he finds that he can't pass Mathematical Analysis! What a sad thing it is! But luckily, he still pass General Physics, so he decides to treat his N friends some Coca-Cola. The Coca-Cola is cheap - it only costs 3 yuan to buy one bottle.

What's more, he knows that the Cola company is holding a promotional activity again that customers can exchange one bottle of cola withM caps. And, because his friends are all good guys, they will give him some caps after drinking the Cola. After he deliberate for a long time, he thinks exactly one Coca-Cola for each person is enough. He is so happy!

But it's not so easy. The Cola company prints some signs on the Cola which is exchanged by caps, such as "Free!", "Gifts!", "Let's black someone!", and so on. We can call it Gift Cola. His friends have different attitude toward to different Cola. To more specific, if Edwardgives the i^th friend a bottle of Normal Cola, he can get Pi caps back; and if he gives a bottle of Gift Cola, he can get Qi caps back.

Edward loves caps and he wants to get the most caps he can! However, the Cola Company often changes the variable M. So Edward asks you how many caps he can get at most in the given M after he treats all his friends.

Remember, Edward is GSF, so he has enough money to buy N bottles of cola, But he can not take the caps and then give the Cola to his friends; Moreover, he can not buy the Cola for himself, too. And he can borrow unlimited caps from someone (such as Dai), too. Of course, he needs to return all the caps he has borrowed at last.

Input

The input contains no more than 30 test cases. Please notice that there's no empty line between each test cases.

For each test case, first line has two integers N (1 ≤ N ≤ 100000) - the number of his friends and T (1 ≤ T ≤ 10000) - the number of the queries Edward asks.

The following N lines, each line contains two integers Pi and Qi (0 ≤ Pi, Qi ≤ 1000), which shows the i^th friend will give back Pi caps forNormal Cola and Qi caps for Gift Cola.

The following T lines, each line contains an integer M (1 ≤ M ≤ 1000), which means Edward can use M caps to exchange for one Gift Cola.

Process to END_OF_FILE.

Output

For each test case, you should output T lines, each line has an integer - the maximal number of caps Edward can get after he treats all his friends.

Sample Input

2 1
2 0
0 2
2

Sample Output

2

题意：你给第i个人喝普通可乐他能给你Pi个盖子，给他中奖可乐他能给你Qi个盖子。中奖可乐用M个盖子可以换一个，盖子可以先借用再去换可乐。问：最多能获得盖子数。

 

思路：按Qi-Pi差值排序。临界点是Qi-Pi=M，这时候换不换中奖可乐，最后得到的盖子是一样的。所以二分下临界点的位置。然后答案就是临界点之前的Pi求和+临界点之后Qi-M求和。

AC代码：

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cstdlib>
 using namespace std;
 struct node
 {
     int pi,qi;
     int cha;
 };

 int pisum[+]={};
 int qisum[+]={};

 bool cmp(node a,node b)
 {
     return a.cha>b.cha;
 }

 node kiss[+];

 int main()
 {
 //    freopen("input.txt","r",stdin);
     int n,T;
     while(scanf("%d%d",&n,&T)==){
         for(int i=;i<n;i++){
             scanf("%d%d",&kiss[i].pi,&kiss[i].qi);
             kiss[i].cha=kiss[i].qi-kiss[i].pi;
         }
         sort(kiss,kiss+n,cmp);
         pisum[]=kiss[].pi;
         qisum[]=kiss[].qi;
         for(int i=;i<n;i++){
             pisum[i]=kiss[i].pi+pisum[i-];//分别求和
             qisum[i]=kiss[i].qi+qisum[i-];
         }
         while(T){
             int m;
             scanf("%d",&m);
             int l=,r=n-;
             int ans=-;
             while(l<=r){//二分查找分界点
                 int ss=(l+r)/;
                 if(kiss[ss].cha<m){
                     r=ss-;
                 }
                 else{
                     ans=ss;
                     l=ss+;
                     if(kiss[ss].cha==m){
                         break;
                     }
                 }
             }
             if(ans==-){//判断输出
                 printf("%d\n",pisum[n-]);
             }
             else{
                 printf("%d\n",qisum[ans]+pisum[n-]-pisum[ans]-(ans+)*m);
             }
             T--;
         }
     }
     return ;
 }