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  • 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 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 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 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 nn and mm (1n,m21051≤n,m≤2⋅105) — the length of aa and the length of bb correspondingly.

    The second line of the input contains one huge integer 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 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
    ");
        else if(M == 1 && A[N - 1] == '0')
            printf("0
    ");
        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
    ", ans % 998244353);
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zlrrrr/p/9822809.html
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