Problem Link : http://www.lightoj.com/volume_showproblem.php?problem=1080
This problem can be solved by using both BIT (Binary Indexed Tree) and Segment tree. Both solution is Here..
/*BIT (Binary Indexed Tree)*/ /* +-+ +-+ +-+ |R| |.| |S| +-+ +-+ +-+ */ #include <bits/stdc++.h> #define pii pair <int,int> #define sc scanf #define pf printf #define Pi 2*acos(0.0) #define ms(a,b) memset(a, b, sizeof(a)) #define pb(a) push_back(a) #define MP make_pair #define db double #define ll long long #define EPS 10E-10 #define ff first #define ss second #define sqr(x) (x)*(x) #define D(x) cout<<#x " = "<<(x)<<endl #define VI vector <int> #define DBG pf("Hi\n") #define MOD 100007 #define MAX 100010 #define CIN ios_base::sync_with_stdio(0); cin.tie(0) #define SZ(a) (int)a.size() #define sf(a) scanf("%d",&a) #define sff(a,b) scanf("%d%d",&a,&b) #define sfff(a,b,c) scanf("%d%d%d",&a,&b,&c) #define loop(i,n) for(int i=0;i<n;i++) #define REP(i,a,b) for(int i=a;i<b;i++) #define TEST_CASE(t) for(int z=1;z<=t;z++) #define PRINT_CASE printf("Case %d:\n",z) #define all(a) a.begin(),a.end() #define intlim 2147483648 #define inf 1000000 #define ull unsigned long long using namespace std; int tree[MAX],n; char str[MAX]; void update2(int indx, int v) { while(indx<=n) { tree[indx]+=v; indx += (indx & (-indx)); } } void update(int i, int j) { update2(i,1); update2(j+1,-1); } int query(int indx) { int sum=0; while(indx) { sum+=tree[indx]; indx-=(indx & (-indx)); } return sum; } int main() { ///freopen("in.txt","r",stdin); ///freopen("out.txt","w",stdout); int t; sf(t); TEST_CASE(t) { ms(tree,0); sc("%s",&str[1]); n=strlen(&str[1]); char ch; int m; sf(m); PRINT_CASE; while(m--) { getchar(); sc("%c",&ch); if(ch=='I') { int a,b; sff(a,b); update(a,b); } else { int a; sf(a); if(query(a)%2==1) { if(str[a]=='0') pf("1\n"); else pf("0\n"); } else pf("%c\n",str[a]); } } } return 0; }
Another solution using segment tree is here -> 1080 – Binary Simulation(Segment Tree)
Can you please , Explain the algorithm behind the coding(BIT Solution)?
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I think you know that BIT works as a dynamic cumulative sum array. Here If an interval i to j is inverted then we are adding 1 with all the index from i to j. after several invert operation if a query operation come. Then we can check the value of that index. if the value is an even number then we know that even number of invert make the initial value so we print the actual value otherwise we print the inverted value.
And BIT works as cumulative sum. So if we add 1 to index i then it means we are adding 1 from i to nth index of the cumulative array. but we need to add 1 only i th to j th index. So we are adding a -1 at j+1 th index. So that extra added one from j+1 th index to nth index will vanish.
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