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Light OJ 1080 - Binary Simulation

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题目链接:http://www.lightoj.com/volume_showproblem.php?problem=1080


1080 - Binary Simulation
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problem=1080" style="color:rgb(79,107,114)">Statistics

Forum
Time Limit: 2 second(s) Memory Limit: 64 MB

Given a binary number, we are about to do some operations on the number. Two types of operations can be here.

‘I i j‘    which means invert the bit from i to j (inclusive)

‘Q i‘    answer whether the ith bit is 0 or 1

The MSB (most significant bit) is the first bit (i.e. i=1). The binary number can contain leading zeroes.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a line containing a binary integer having length n (1 ≤ n ≤ 105). The next line will contain an integer q (1 ≤ q ≤ 50000) denoting the number of queries. Each query will be either in the form ‘I i j‘where i, j are integers and 1 ≤ i ≤ j ≤ n. Or the query will be in the form ‘Q i‘ where i is an integer and 1 ≤ i ≤ n.

Output

For each case, print the case number in a single line. Then for each query ‘Q i‘ you have to print 1 or 0 depending on the ith bit.

Sample Input

Output for Sample Input

2

0011001100

6

I 1 10

I 2 7

Q 2

Q 1

Q 7

Q 5

1011110111

6

I 1 10

I 2 7

Q 2

Q 1

Q 7

Q 5

Case 1:

0

1

1

0

Case 2:

0

0

0

1

Note

Dataset is huge, use faster i/o methods.


PROBLEM SETTER: JANE ALAM JAN

题目大意:给你一段二进制数,当输入为 I  x  y时。就把这段变成它的反串 。当为Q 就查询当前x这个位置的数字视为0还是1。


思路:此题有两种解题的思路,一种是线段树。一种是树状数组。当用树状数组时。能够看成是一个区间改动,单点查询的问题,当我们用线段树时。那么就须要用到懒惰标记。

对于这道题树状数组确实比线段是快非常多。

。。。


线段树代码:

//time 584 ms

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#define L(u) u<<1
#define R(u) u<<1|1

using namespace std;

const int Max=100010;
char str[Max];
class Node
{
public:
    int l,r;
    char str;
    int add;
}node[Max*4];

void build(int l,int r,int u)
{
    node[u].l=l;
    node[u].r=r;
    node[u].add=0;
    if(l==r)
    {
        node[u].str=str[l];
        return;
    }
    int mid=(l+r)/2;
    build(l,mid,L(u));
    build(mid+1,r,R(u));
}

void pushdown(int u)                                   //注意这儿要懒惰标记
{
    node[L(u)].add+=node[u].add;                       //主义要加上。。。

。 node[R(u)].add+=node[u].add; node[u].add=0; } void update(int l,int r,int u) { if(l<=node[u].l&&node[u].r<=r) { node[u].add++; return ; } if(node[u].add) pushdown(u); int mid=(node[u].l+node[u].r)/2; if(r<=mid) { update(l,r,L(u)); } else if(l>mid) { update(l,r,R(u)); } else { update(l,mid,L(u)); update(mid+1,r,R(u)); } return ; } char query(int l,int r,int u) { if(l<=node[u].l&&node[u].r<=r) { if((node[u].add%2)==1) { node[u].add=0; if(node[u].str=='1') { node[u].str='0'; return node[u].str; } else { node[u].str='1'; return node[u].str; } } else { node[u].add=0; return node[u].str; } } if(node[u].add) pushdown(u); int mid=(node[u].l+node[u].r)/2; if(r<=mid) { return query(l,r,L(u)); } else if(l>mid) { return query(l,r,R(u)); } } int main() { int T,q,i,j,kk=0;; scanf("%d",&T); while(T--) { getchar(); kk++; str[0]='3'; scanf("%s",str+1); int len=strlen(str); build(1,len-1,1); scanf("%d",&q); char ch[5]; printf("Case %d:\n",kk); for(i=0;i<q;i++) { scanf("%s",ch); if(ch[0]=='I') { int x,y; scanf("%d%d",&x,&y); update(x,y,1); } else { int x; scanf("%d",&x); printf("%c\n",query(x,x,1)); } } } return 0; }


树状数组代码:

//time 304 ms
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>

using namespace std;

const int Max=100010;

int a[Max];
int n;

int lowbit(int x)
{
    return x&(-x);
}

void insert(int x,int d)
{
    while(x<=n)
    {
        a[x]+=d;
        x+=lowbit(x);
    }
}

int query(int x)
{
    int res=0;
    while(x>0)
    {
        res+=a[x];
        x-=lowbit(x);
    }
    return res;
}

int main()
{
    char str[Max],ch[5];
    int T,q,kk=0;
    scanf("%d",&T);
    while(T--)
    {
        kk++;
        memset(a,0,sizeof(a));
        str[0]='2';
        scanf("%s",str+1);
        n=strlen(str)-1;
        scanf("%d",&q);
        printf("Case %d:\n",kk);
        for(int i=0;i<q;i++)
        {
            scanf("%s",ch);
            if(ch[0]=='I')
            {
                int x,y;
                scanf("%d%d",&x,&y);
                insert(x,1);
                insert(y+1,-1);
            }
            else
            {
                int x;
                scanf("%d",&x);
                int t=query(x);
                if(t%2==1)
                {
                    if(str[x]=='1')
                    {
                        printf("0\n");
                    }
                    else
                    {
                        printf("1\n");
                    }
                }
                else
                {
                    printf("%c\n",str[x]);
                }
            }
        }

    }
    return 0;
}













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