UVA

http://acm.bnu.edu.cn/v3/problem_show.php?pid=27853

Interval Product

Time Limit: 2000ms

Memory Limit: 131072KB

This problem will be judged on UVA. Original ID: 12532
64-bit integer IO format: %lld Java class name: Main

Prev

Submit Status Statistics Discuss

Next

Type:

None

 

None Graph Theory 2-SAT Articulation/Bridge/Biconnected Component Cycles/Topological Sorting/Strongly Connected Component Shortest Path Bellman Ford Dijkstra/Floyd Warshall Euler Trail/Circuit Heavy-Light Decomposition Minimum Spanning Tree Stable Marriage Problem Trees Directed Minimum Spanning Tree Flow/Matching Graph Matching Bipartite Matching Hopcroft–Karp Bipartite Matching Weighted Bipartite Matching/Hungarian Algorithm Flow Max Flow/Min Cut Min Cost Max Flow DFS-like Backtracking with Pruning/Branch and Bound Basic Recursion IDA* Search Parsing/Grammar Breadth First Search/Depth First Search Advanced Search Techniques Binary Search/Bisection Ternary Search Geometry Basic Geometry Computational Geometry Convex Hull Pick's Theorem Game Theory Green Hackenbush/Colon Principle/Fusion Principle Nim Sprague-Grundy Number Matrix Gaussian Elimination Matrix Exponentiation Data Structures Basic Data Structures Binary Indexed Tree Binary Search Tree Hashing Orthogonal Range Search Range Minimum Query/Lowest Common Ancestor Segment Tree/Interval Tree Trie Tree Sorting Disjoint Set String Aho Corasick Knuth-Morris-Pratt Suffix Array/Suffix Tree Math Basic Math Big Integer Arithmetic Number Theory Chinese Remainder Theorem Extended Euclid Inclusion/Exclusion Modular Arithmetic Combinatorics Group Theory/Burnside's lemma Counting Probability/Expected Value Others Tricky Hardest Unusual Brute Force Implementation Constructive Algorithms Two Pointer Bitmask Beginner Discrete Logarithm/Shank's Baby-step Giant-step Algorithm Greedy Divide and Conquer Dynamic Programming Tag it!

[PDF Link]

It's normal to feel worried and tense the day before a programming contest. To relax, you went out for a drink with some friends in a nearby pub. To keep your mind sharp for the next day, you decided to play the following game. To start, your friends will give you a sequence of N integers X₁, X₂,..., XN. Then, there will be K rounds; at each round, your friends will issue a command, which can be:

• a change command, when your friends want to change one of the values in the sequence; or
• a product command, when your friends give you two values I, J and ask you if the product X_I x X_I+1 x ... x X_J-1 x X_J is positive, negative or zero.

Since you are at a pub, it was decided that the penalty for a wrong answer is to drink a pint of beer. You are worried this could affect you negatively at the next day's contest, and you don't want to check if Ballmer's peak theory is correct. Fortunately, your friends gave you the right to use your notebook. Since you trust more your coding skills than your math, you decided to write a program to help you in the game.

Input

Each test case is described using several lines. The first line contains two integers N and K, indicating respectively the number of elements in the sequence and the number of rounds of the game ( 1N, K10⁵). The second line contains N integers X_i that represent the initial values of the sequence ( -100X_i100 for i = 1, 2,..., N). Each of the next K lines describes a command and starts with an uppercase letter that is either `C' or `P'. If the letter is `C', the line describes a change command, and the letter is followed by two integers I and V indicating that X<sub>I</sub> must receive the value V ( 1IN and -100V100). If the letter is `P', the line describes a product command, and the letter is followed by two integers I and J indicating that the product from X_I to X_J, inclusive must be calculated ( 1IJN). Within each test case there is at least one product command.

Output

For each test case output a line with a string representing the result of all the product commands in the test case. The i-th character of the string represents the result of the i-th product command. If the result of the command is positive the character must be `+' (plus); if the result is negative the character must be `-' (minus); if the result is zero the character must be `0' (zero).

Sample Input

4 6
-2 6 0 -1
C 1 10
P 1 4
C 3 7
P 2 2
C 4 -5
P 1 4
5 9
1 5 -2 4 3
P 1 2
P 1 5
C 4 -5
P 1 5
P 4 5
C 3 0
P 1 5
C 4 -5
C 4 -5

Sample Output

0+-
+-+-0

#include<iostream>
#include<cstdio>
#define maxn 1000005
struct  node
{
    int left,right;
    int num;
}tree[3*maxn];
int out=1;
void build(int left,int right,int i)
{
    tree[i].left=left;
    tree[i].right=right;
    tree[i].num=0;
    if(tree[i].left==tree[i].right)
         return ;
    int mid=(left+right)/2;
    build(left,mid,2*i);
    build(mid+1,right,2*i+1);
}
void insert(int id,int i,int j)
{
      if(tree[id].left==i&&tree[id].right==i)
               {

                   tree[id].num=j;
                }
      if(tree[id].left==tree[id].right)
                 return ;
    if(i>tree[id].right)
        return;
    if(i<tree[id].left)
            return;
     int mid=(tree[id].left+tree[id].right)/2;
     if(i<=mid)
         insert(id*2,i,j);
     else
         insert(id*2+1,i,j);
    tree[id].num=tree[2*id].num*tree[2*id+1].num;

}
void sum(int id,int i,int j)
{
    int mid=(tree[id].left+tree[id].right)/2;
    if(tree[id].left==i&&tree[id].right==j)
     {

         out*=tree[id].num;
             return ;
     }
     if(j<=mid)
       sum(id*2,i,j);
     else if(i>mid)
         sum(id*2+1,i,j);
     else
     {
        sum(id*2,i,mid);
        sum(id*2+1,mid+1,j);
     }
}
int main()
{
    int i,x,y,n,m,u,num;

   while(~ scanf("%d%d",&n,&m))
   {
       char str[1000006];
       build(1,n,1);
     for(i=1;i<=n;i++)
    {
        scanf("%d",&u);
        if(u>0)
           u=1;
         else if(u==0)
               u=0;
         else
           u=-1;
      //  printf("u=%d
",u);
        insert(1,i,u);

    }

    char p;
    int r=0;
    for(i=1;i<=m;i++)
    {
     // printf("AAA
");
        getchar();
       scanf("%c",&p);
        //printf("p=%c
",p);
          if(p=='C')
             {
                scanf("%d%d",&x,&y);
             if(y>0)
               y=1;
            else if(y==0)
               y=0;
            else
               y=-1;
           // printf("y=%d
",y);
                 //printf("aa
");
                 insert(1,x,y);
              }
          if(p=='P')
            {
                scanf("%d%d",&x,&y);
                sum(1,x,y);
              //  printf("out=%d
",out);
                 if(out>0)
                  {
                     str[r++]='+';
                    //  printf("+
");
                  }
                 else if(out<0)
                     {
                         str[r++]='-';
                         // printf("-
");
                     }
                  else
                  {
                     str[r++]='0';
                     // printf("0
");
                  }
                        out=1;
            }
        }
        str[r] = '';
        printf("%s
",str);
   }
     return 0;
}