Xor and Sum

                            Solution in C :

In C++ :





#include <iostream>
#include <algorithm>
#include <vector>
#include <string>

using namespace std;

#define long long long
#define mod 1000000007ll
#define M 500500
#define N 13001000


const int T = 314159;

string a, b;
long p[M], x[M], s[N];

void pre() {
    p[0] = 1;
    for (int i = 1; i < M; ++i)
        p[i] = (2 * p[i - 1]) % mod;
}

void read() {
    cin >> a >> b;

    for (int i = 0; i < (int) a.length(); ++i)
        x[a.length() - 1 - i] = a[i] == '1' ? 1 : 0;

    for (int i = 0; i < (int) b.length(); ++i)
        s[b.length() - 1 - i + M] = b[i] == '1' ? 1 : 0;

    for (int i = 1; i < N; ++i)
        s[i] += s[i - 1];
}

long sum(int l, int r) {
    return s[r + M] - s[l + M];
}

void kill() {
    long ans = 0;
    for (int i = 0; i < M; ++i)
    if (x[i] == 0)
        ans = (ans + p[i] * sum(i - T - 1, i)) % mod;
    else
        ans = (ans + p[i] * (T + 1 - sum(i - T - 1, i))) % mod;

    cout << ans << "\n";
}


int main() {
#ifdef TROLL
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
#else
    ios_base::sync_with_stdio(0);
#endif

    pre();
    read();
    kill();

    return 0;
}








In Java :





import java.util.Scanner;

public class Solution {
	public static void main(String[] args) {
		final int n = 314159; 
		final int maxlen = 1000000;
		final int mod = 1000000007;
		Scanner in = new Scanner(System.in);
		char a[] = in.nextLine().toCharArray();
		char b[] = in.nextLine().toCharArray();
		int bina[] = new int[maxlen];
		int binb[] = new int[maxlen];
		int bone[] = new int[maxlen];
		for(int i=0;i<a.length;i++){
			bina[i]=a[a.length-1-i]-'0';
		}
		for(int i=a.length;i<maxlen;i++){
			bina[i]=0;
		}
		for(int i=0;i<b.length;i++){
			binb[i]=b[b.length-1-i]-'0';
		}
		for(int i=b.length;i<maxlen;i++){
			binb[i]=0;
		}
		bone[0]=binb[0];
		for(int i = 1;i <= n;i++){
			bone[i]=bone[i-1]+binb[i];
		}
		for(int i=n+1;i<1000000;i++){
			bone[i]=bone[i-1]+binb[i]-binb[i-n-1];
		}
		long sum = 0;
		long mul = 1;
		for(int i=0;i<maxlen;i++){
			if(bina[i]==1){
				sum = (sum + (mul*(n+1-bone[i]))%mod)%mod;	
			} else if(bina[i]==0){
				sum = (sum + (mul*(bone[i]))%mod)%mod;
			}
			mul=(mul*2)%mod;
		}
		System.out.println(sum);
	}
}








In C :






#include <stdio.h>
#include <stdlib.h> 
#define MOD 1000000007
#define s(k) scanf("%d",&k);
#define sll(k) scanf("%lld",&k);
#define p(k) printf("%d\n",k);
#define pll(k) printf("%lld\n",k);
#define f(i,N) for(i=0;i<N;i++)
#define f1(i,N) for(i=0;i<=N;i++)
#define f2(i,N) for(i=1;i<=N;i++)
#define lim 314160
typedef long long ll;
void rev(char* in){
    int start=0,end=strlen(in)-1,i,l=strlen(in);
    char t;
    while(start<end){
        t=in[start];
        in[start]=in[end];
        in[end]=t;
        start++;
        end--;
    }
    for(i=strlen(in);i<lim+l;i++)
        in[i]='0';
}
int main(){
    char A[414160],B[414160];
    int i,num,len;
    ll X,sum,pos;
    scanf("%s",A);
    scanf("%s",B);
    len=strlen(B);
    rev(A);
    rev(B);
    num=0;
    pos=1;
    sum=0;
    for(i=0;i<lim-1;i++){
        if(B[i]=='1')
            num++;
        X=num;
        if(A[i]=='1')
            X=lim-X;
        sum=(sum+X*pos)%MOD;
        pos=(pos<<1)%MOD;
    }
    for(i=0;i<len;i++){
        X=num;
        sum=(sum+X*pos)%MOD;
        pos=(pos<<1)%MOD;
        if(B[i]=='1')
            num--;
    }
    
    pll(sum);
/*    printf("%s\n",A);
    printf("%s\n",B);*/
    return 0;
    
}








In Python3 :






a = int( input( ), 2 )
b = int( input( ), 2 )
r = 0

for i in range( 314160 ):
    r += a ^ b
    b *= 2

print( (r % (10**9+7)) )