CodeForces Round #515 Div.3 E. Binary Numbers AND Sum

http://codeforces.com/contest/1066/problem/E

 

You are given two huge binary integer numbers $\mathrm{aa and bb of\; lengths nn and mm respectively.\; You\; will\; repeat\; the\; following\; process:\; if b>0b>0,\; then\; add\; to\; the\; answer\; the\; value a \& ba \& b and\; divide bb by 22 rounding\; down\; (i.e.\; remove\; the\; last\; digit\; of bb),\; and\; repeat\; the\; process\; again,\; otherwise\; stop\; the\; process.}$

The value $\mathrm{a \& ba \& b means\; bitwise AND of aa and bb.\; Your\; task\; is\; to\; calculate\; the\; answer\; modulo 998244353998244353.}$

Note that you should add the value $\mathrm{a \& ba \& b to\; the\; answer\; in\; decimal\; notation,\; not\; in\; binary.\; So\; your\; task\; is\; to\; calculate\; the\; answer\; in\; decimal\; notation.\; For\; example,\; if a=10102 (1010)a=10102 (1010) and b=10002 (810)b=10002 (810),\; then\; the\; value a \& ba \& b will\; be\; equal\; to 88,\; not\; to 10001000.}$

Input

The first line of the input contains two integers $\mathrm{nn and mm (1\le n,m\le 2\cdot 1051\le n,m\le 2\cdot 105)\; —\; the\; length\; of aa and\; the\; length\; of bb correspondingly.}$

The second line of the input contains one huge integer $\mathrm{aa.\; It\; is\; guaranteed\; that\; this\; number\; consists\; of\; exactly nn zeroes\; and\; ones\; and\; the\; first\; digit\; is\; always 11.}$

The third line of the input contains one huge integer $\mathrm{bb.\; It\; is\; guaranteed\; that\; this\; number\; consists\; of\; exactly mm zeroes\; and\; ones\; and\; the\; first\; digit\; is\; always 11.}$

Output

Print the answer to this problem in decimal notation modulo $998244353998244353.$

Examples

input

Copy

4 4
1010
1101

output

Copy

12

input

Copy

4 5
1001
10101

output

Copy

11

代码：

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
int N, M;
char A[200010], B[200010];
ll sum[200010];

ll Pow(ll x, ll n, ll mod) {
    ll res = 1;
    while(n > 0) {
        if(n % 2 == 1) {
            res = res * x;
            res = res % mod;
        }
        x = x * x;
        x = x % mod;
        n >>= 1;
    }
    return res;
}

int main() {
    scanf("%d%d", &N, &M);
    scanf("%s%s", A, B);
    memset(sum, 0, sizeof(sum));
    for(int i = N - 1; i >= 0; i --) {
        sum[N - 1 - i] =(A[i] - '0') * Pow(2, N - 1 - i, 998244353) + sum[N - i - 2];
        sum[N - 1 - i] %= 998244353;
    }

    if(M == 1 && A[N - 1] == '1')
        printf("1\n");
    else if(M == 1 && A[N - 1] == '0')
        printf("0\n");
    else {
        ll ans = 0;
        for(int i = 0; i <= N / 2 - 1; i ++)
            swap(A[i], A[N - 1 - i]);
        for(int i = 0; i <= M / 2 - 1; i ++)
            swap(B[i], B[M - 1 - i]);

        for(int i = 0; i < M; i ++) {
            if(B[i] == '1') {
                if(i >= N)
                    sum[i] = sum[N - 1];
                ans += sum[i];
                ans %= 998244353;
            }
        }
        printf("%I64d\n", ans % 998244353);
    }
    return 0;
}