hnoj beads

http://acm.hnu.cn/online/?action=problem&type=show&id=12515

Beads

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65366KB

Total submit users: 33, Accepted users: 29

Problem 12515 : No special judgement

Problem description

A game consists of putting beads in boxes. The rules of the game are too complex to describe here, but all you need to know is that keeping track of the number of beans in adjacent boxes are very important to the outcome of the game. You are asked by a friend to write a program to help him win the game every time. At the start of a game, all boxes are empty.

Input

The first line of the input consists of a single number T, the number of games played. Each game start with a line describing B, P and Q, the number of boxes, put requests and query requests, respectively. Then follows P + Q lines with either P i a, saying a beads are put in box number i,or Q i j, a query request for the number of beads in boxes i through j, inclusive.

Output

For each query request, output the number of beads in boxes a through b, inclusive, that are in the boxes at this point of the game.

Sample Input

1 7 5 3 P 2 1 P 3 3 P 4 7 Q 1 4 P 7 6 Q 4 4 P 6 4 Q 1 7

Sample Output

11 7 21

Judge Tips

 0 < T  100  0 < B  100000  0 < P  30000  0 < Q <= 30000  0 <= a <= 100  0 < i <= j <= B  Note that boxes are 1-indexed.  This is an I/O-heavy problem. For Java programmers, this means that you should use BufferedReader for input reading (not Scanner). It is also bene cial to build all output in a StringBuilder before printing in a single print statement.

Problem Source

IDIOPEN 2011

直接进行树状数组的调用就好了~~ 用g++过了~~~用C就过不了~~坑啊

post code：

#include<stdio.h>
#include<string.h>
int a[1010000],b;

int lowbit(int t)   //lowbit的进制转换 查找二的指数形式
{
     return t&(t^(t-1));   
}

void insert(int pos,int num)  //插入数据更新pos到b的节点的数据的值
{
     while(pos<=b)
     {
       a[pos]+=num;
       pos+=lowbit(pos);                         
     }
}

int query(int end)    //记录1到end的和
{
   int sum=0;
   while( end>0 )
   {
     sum+=a[end];
     end-=lowbit(end);       
   }    
   return sum;
}

int main()
{
      int x,p,q,i,pos,num,beg,end,sumbeg,sumend;
      char judge[3];
      scanf("%d",&x);
      while(x--)
      {
           memset( a, 0, sizeof(a) );
           scanf("%d %d %d",&b,&p,&q);
           for(i=1;i<=p+q;i++)
              {
              scanf("%s",judge);
              if( judge[0]=='P' ){ scanf("%d %d",&pos,&num); insert(pos,num); }
              if( judge[0]=='Q' ){
                                   scanf("%d %d",&beg,&end);  //这里要注意 要求beg到end的和 需要用query（end）-query（beg-1）来求解
                                   
                                   sumbeg=query(beg-1);
                                   sumend=query(end);
                                   printf("%d\n",sumend-sumbeg);
                                 } 
           
              }



                
      }      
    
}