Problem Link : http://www.spoj.com/problems/SUMSUM/
Solution Idea:
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This one is a interesting problem. The main idea behind the problem is combinatorics. At first we need to count the number of 1 bit and number of 0 bit in the i_th position of a given number. Then we need to do some combinatorial calculation.
Now think about an array of 4 number A=[1,0,1,0]. If we perform OR operation in the range 1 to 4 what will we get? We get 5 as answer. If we form pair from this number and perform OR operation between then then we will get the answer. Now as here is 4 element so we can perform pair in 4*(4-1)/2 = 6 ways. Now comes the tricky part. For OR operation how many pair will contribute nothing ?? Here is 2 zero and we can form 1 pair from this two element who will not contribute anything. So for i_th bit position the contribution for OR operation is number_of_active_pair*(2^i).
For AND operation we need to consider the number of pair can be form by 1 bit and for XOR the number of pair will be number_of_one_bit*number_of_zero_bit.
We can consider a number as an array of 30 element whose value is either 0 or 1 and perform above operation.
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pii pair <int,int>
#define pll pair <long long,long long>
#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) cerr<<#x " = "<<(x)<<endl
#define VI vector <int>
#define DBG pf("Hi\n")
#define MOD 1000000007
#define CIN ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a) (int)a.size()
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v) for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n) for(int i=0;i<n;i++)
#define loop1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define RREP(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: ",z)
#define LINE_PRINT_CASE printf("Case %d:\n",z)
#define CASE_PRINT cout<<"Case "<<z<<": "
#define all(a) a.begin(),a.end()
#define intlim 2147483648
#define infinity (1<<28)
#define ull unsigned long long
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
using namespace std;
//using namespace __gnu_pbds;
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
//const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move
//const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move
/*------------------------------------------------*/
/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/
#define mx 100005
int tree[28][mx];
void update(int id, int idx, int val)
{
for(;idx<mx && idx;idx+=idx&-idx)
tree[id][idx]+=val;
}
ll query(int id, int idx)
{
ll ret=0;
for(;idx;idx-=idx&-idx)
ret+=tree[id][idx];
return ret;
}
int ara[mx];
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int n,q;
sff(n,q);
for(int i=1;i<=n;i++) sf(ara[i]);
for(int i=1;i<=n;i++)
{
for(int j=0;j<28;j++)
{
if(check(ara[i],j))
update(j,i,1);
}
}
while(q--)
{
int a,b,c;
sf(a);
if(a==1)
{
sff(b,c);
for(int i=0;i<28;i++)
{
if(check(ara,i))
update(i,c,-1);
}
ara=b;
for(int i=0;i<28;i++)
{
if(check(ara,i))
update(i,c,1);
}
}
else
{
char ss[10];
scanf(" %s",&ss);
string str=string(ss);
sff(b,c);
ll temp[30];
for(int i=0;i<28;i++)
{
temp[i]=query(i,c);
}
for(int i=0;i<28;i++)
{
temp[i]-=query(i,b-1);
}
ll ans=0;
for(int i=0;i<28;i++)
{
ll one=temp[i];
ll zero=(c-b+1)-one;
ll pairs=0;
if(str=="OR")
{
ll total=one+zero;
pairs=(total*(total-1))/2;
pairs-=(zero*(zero-1))/2;
}
else if(str=="AND")
{
pairs=(one*(one-1))/2;
}
else if(str=="XOR")
{
pairs=one*zero;
}
ans+=(1LL<<i)*pairs;
}
printf("%lld\n",ans);
}
}
return 0;
}