Problem:
Score : 200 points

Problem Statement

Given are strings S and T. Consider changing S to T by repeating the operation below. Find the minimum number of operations required to do so.

Operation: Choose one character of S and replace it with a different character.

Constraints

S and T have lengths between 1 and 2\times 10^5 (inclusive).

S and T consists of lowercase English letters.

S and T have equal lengths.

Input

Input is given from Standard Input in the following format:

S
T

Output

Print the answer.

Sample Input 1

cupofcoffee
cupofhottea

Sample Output 1

4

We can achieve the objective in four operations, such as the following:

First, replace the sixth character c with h.

Second, replace the eighth character f with t.

Third, replace the ninth character f with t.

Fourth, replace the eleventh character e with a.

Sample Input 2

abcde
bcdea

Sample Output 2

5

Sample Input 3

apple
apple

Sample Output 3

0

No operations may be needed to achieve the objective.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <queue>

#define rep(i,n) for(int i = 0; i < (n); ++i)
#define MOOD 1000000007


const double PI=3.14159265358979323846;

using namespace std;
using ll = long long;
using P = pair<int,int>;

int cmpnum(const void *a, const void *b)
{
    long long *A = (long long *)a;
    long long *B = (long long *)b;
    if (*A > *B)
        return 1;
    if (*A < *B)
        return -1;
    return 0;
}

int char_sort_max( const void * a , const void * b ) {
    return strcmp(( char * )a , ( char * )b );
}

int char_sort_min( const void * a , const void * b ) {
    return strcmp(( char * )a , ( char * )b );
}

int gcd(int x, int y)
{
    int t;

    while (y != 0) {
        t = x % y;  x = y;  y = t;
    }
    return x;
}/*最大公約数*/

ll lcm(ll a, ll b)
{
    return (ll)((ll)(a * b) / (ll)(gcd(a, b)));
}/*最小公倍数*/

bool IsPrime(int num)
{
    if (num < 2) return false;
    else if (num == 2) return true;
    else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            // 素数ではない
            return false;
        }
    }

    // 素数である
    return true;
}/*素数判定*/


vector<pair<long long, long long> > prime_factorize(long long N) {
    vector<pair<long long, long long> > res;
    for (long long a = 2; a * a <= N; ++a) {
        if (N % a != 0) continue;
        long long ex = 0; // 指数

        // 割れる限り割り続ける
        while (N % a == 0) {
            ++ex;
            N /= a;
        }

        // その結果を push
        res.push_back({a, ex});
    }

    // 最後に残った数について
    if (N != 1) res.push_back({N, 1});
    return res;
}/*素因数分解*/


struct UnionFind{
    vector<int> d;
    UnionFind(int n = 0):d(n,-1){};

    int find(int x){
        if(d[x] < 0) {
            return x;
        }
        else{
            return d[x] = find(d[x]);
        }
    }
    bool unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) return false;
        if (d[x] > d[y]) swap(x, y);

        d[x] += d[y];
        d[y] = x;
        return true;
    }
    bool same(int x, int y){
        return find(x) == find(y);
    }
    int size(int x){
        return -d[find(x)];
    }

};

struct mint {
    ll x;

    mint(ll x) : x((x % MOOD + MOOD) % MOOD) {}

    mint operator-() const { return mint(-x); }

    mint &operator+=(const mint a) {
        if ((x += a.x) >= MOOD) x -= MOOD;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += MOOD-a.x) >= MOOD) x -= MOOD;
        return *this;
    }

    mint &operator*=(const mint a) {
        if ((x *= a.x) %= MOOD);
        return *this;
    }

    mint operator+(const mint a)const {
        mint res(*this);
        return res += a;
    }

    mint operator-(const mint a) const {
        mint res(*this);
        return res -= a;
    }

    mint operator*(const mint a) const {
        mint res(*this);
        return res *= a;
    }

    mint pow(ll t) const{
        if(!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if(t&1) a *= *this;
        return a;
    }

    mint inv() const {
        return pow(MOOD-2);
    }

    mint& operator/=(const mint a) {
        return (*this) *= a.inv();
    }

    mint operator/(const mint a) const {
        mint res(*this);
        return res/=a;
    }

};

mint kaijyo(int n) {
    if (n == 0) return 1;
    mint x = kaijyo(n/2);
    x *= x;
    if (n%2 == 1) x *= 2;
    return x;
}

mint choose(int n, int a) {
    mint x = 1, y = 1;
    rep(i,a) {
        x *= n-i;
        y *= i+1;
    }
    return x / y;
}

int shakutori(ll x, ll n,ll k, ll a[])
{
    long long int ai,res,ok,ng,mid;
    res = 0;
    for (int i = 0; i < n; i++)
    {
        // printf("%lld\n", res);
        ai = a[i];
        if(ai == 0){
            if(x < 0){
                res += 0;
            } else {
                res += n - 1;
            }
        } else
        {
            if(ai <= -1){
                if (ai * ai <= x){
                    res -= 1;
                }
                ok = n;
                ng = -1;
                while (ok-ng>1)
                {
                    mid = (ok + ng) / 2;
                    if(a[mid]*ai<=x){
                        ok = mid;
                    } else {
                        ng = mid;
                    }
                }
                res += n - ok;
            } else
            {
                if (ai * ai <= x)
                {
                    res -= 1;
                }
                ok = -1;
                ng = n;
                while (ng - ok > 1)
                {
                    mid = (ok + ng) / 2;
                    if (a[mid] * ai <= x)
                    {
                        ok = mid;
                    }
                    else
                    {
                        ng = mid;
                    }
                }
                res += ok + 1;
            }

        }
    }
    // printf("%lld\n", res);
    return k <= res / 2;
}

int two_n(int tei, int jyousuu){
    if(jyousuu == 0) return 1;
    int x = two_n(tei,jyousuu/2);
    x *= x ;
    if(jyousuu % 2 == 1) {
        x *= tei;
    }
    return x;




}

ll two_n_long(ll tei, ll jyousuu){
    if(jyousuu == 0) return 1;
    ll x = two_n(tei,jyousuu/2);
    x *= x ;
    if(jyousuu % 2 == 1) {
        x *= tei;
    }
    return x;




}

vector<vector<int>> bit_search(int N){
    vector<vector<int>> bit_sum(two_n(2,N));
    int num = 0;
    for(int bit = 0; bit < (1<<N); ++bit){
        for(int i = 0; i < N; ++i){
            if(bit & (1 << i)){
                bit_sum[num].push_back((i));
            }
        }
        num++;
    }
    return bit_sum;
}


vector<vector<int>> bit_search_array(int N){
    vector<vector<int>> bit_sum(two_n(2,N),vector<int>(N,0));
    int num = 0;
    for(int bit = 0; bit < (1<<N); ++bit){
        for(int i = 0; i < N; ++i){
            if(bit & (1 << i)){
                bit_sum[bit][i] = 1;
            }
        }
        num++;
    }
    return bit_sum;
}




struct jyanken{
    int tensuu;
    char te;
    char aite;
};

int cmp_tensu(const void * n1, const void * n2)
{
    if (((jyanken *)n1)->tensuu < ((jyanken *)n2)->tensuu)
    {
        return 1;
    }
    else if (((jyanken *)n1)->tensuu > ((jyanken *)n2)->tensuu)
    {
        return -1;
    }
    else
    {
        return 0;
    }
}/*structの要素でのソート*/

int main() {


    char S[2000000];
    char T[2000000];
    cin >> S >> T;

    int N = strlen(S);
    int ans = 0;
    for(int i = 0; i < N; ++i){
        if(S[i] != T[i]){ans++;}
    }

    cout << ans;


}


int A(){




    return 0;

}

int B(){
    int N,K;
    cin >> N >> K;
    vector<int> A(N);
    for(int i = 0; i < N; ++i ){
        cin >> A[i];
    }
    sort(A.begin(),A.end());

    int ans = 0;
    for(int i = 0; i < K; ++i ){
        ans += A[i];
    }

    cout << ans;
}

int C(){
    int N;
    cin >> N;
    int ans = 0;
    int now = 0;

    int a[N];
    cin >> a[0];

    for(int i = 1; i < N; ++i){
        cin >> a[i];
        if(a[i] <= a[i-1]){
            ans++;
        }
        else{
            if(now < ans){
                now = ans;
            }
            ans = 0;
        }

    }

    cout << max(now,ans);




    return 0;
}

int D(){
    ll N,Q;
    ll ans = 0;
    cin >> N;
    vector<ll> A(100001,0);
    for(int i = 0; i < N; ++i ){
        int a;
        cin >> a;
        A[a]++;
        ans += a;
    }

    cin >> Q;
    for(int i = 0; i < Q; ++i){

        int b,c;
        cin >> b >> c;
        A[c] += A[b];
        ans +=   A[b] * (c - b);
        A[b] = 0;

        cout << ans << '\n';

    }


    return 0;
}