二進制 set public std images pos mos it is src

題目鏈接：http://www.lightoj.com/volume_showproblem.php?problem=1080

1080 - Binary Simulation

PDF (English)

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

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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 i^th 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 i^th 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

題目大意：給你一段二進制數，當輸入為 I x y時。就把這段變成它的反串 。當為Q 就查詢當前x這個位置的數字視為0還是1。

思路：此題有兩種解題的思路，一種是線段樹。一種是樹狀數組。當用樹狀數組時。能夠看成是一個區間改動，單點查詢的問題，當我們用線段樹時。那麽就須要用到懶惰標記。

對於這道題樹狀數組確實比線段是快非常多。

。。。

線段樹代碼：

//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)                                   //註意這兒要懶惰標記
{
    node[L(u)].add+=node[u].add;                       //主義要加上。。。
。
。
    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;
}

樹狀數組代碼：

//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;
}

Light OJ 1080 - Binary Simulation