Informatics Olympiad One Pass 2.1: High-precision calculation

​Part 2 Basic Algorithm

Chapter One High-precision Calculation

1307 High-precision multiplication

#include <iostream>
using namespace std;

const int N = 1e5 + 10;
int a[N], b[N], c[N], x;

void init(int a[]) {
    string s;
    cin >> s;

    a[0] = s.size();
    for (int i = 1; i <= a[0]; i ++ ) {
        a[i] = s[a[0]-i] - '0';
    }
}

void dy(int a[]) {
    for (int i = a[0]; i >= 1; i --) {
        cout << a[i];
    }

    cout << endl;
}

//高精度 a * 高精度 b = 高精度 c
void mul2(int a[], int b[], int c[]) {
    for (int i = 1; i <= a[0]; i ++ ) {
        int t = 0;
        for (int j = 1; j <= b[0]; j ++ ) {
            t += c[i+j-1] + a[i]*b[j];
            c[i+j-1] = t % 10;
            t /= 10;
        }
        if (t) c[i+b[0]] = t;
    }

    c[0] = a[0] + b[0];
    while(!c[c[0]] && c[0] > 1) c[0] -- ;
}

int main() {
    init(a);
    init(b);

    mul2(a, b, c);

    dy(c);

    return 0;
}

1308 High Precision

#include <iostream>
using namespace std;

const int N = 1e5 + 10;
int a[N], b[N], c[N], x;

void init(int a[]) {
    string s;
    cin >> s;

    a[0] = s.size();
    for (int i = 1; i <= a[0]; i ++ ) {
        a[i] = s[a[0]-i] - '0';
    }
}

void dy(int a[]) {
    for (int i = a[0]; i >= 1; i --) {
        cout << a[i];
    }

    cout << endl;
}

//高精度 a - 高精度 b = 高精度 c
void sub(int a[], int b[], int c[]) {
    int t = 0;
    c[0] = a[0];

    for (int i = 1; i <= c[0]; i ++ ) {
        t = a[i] - b[i] - t;
        c[i] = (t + 10) % 10;
        t = t < 0 ? 1 : 0;
    }
    while (!c[c[0]] && c[0] > 1) c[0] -- ;
}

//如果高精度 a > 高精度 b，返回 1
//如果高精度 a < 高精度 b，返回 -1
//如果高精度 a = 高精度 b，返回 0
int cmp(int a[], int b[]) {
    while (!a[a[0]] && a[0] > 1) a[0] -- ;
    while (!b[b[0]] && b[0] > 1) b[0] -- ;

    if (a[0] > b[0]) return 1;
    if (a[0] < b[0]) return -1;
    for (int i = a[0]; i >= 1; i -- ) {
        if (a[i] > b[i]) return 1;
        if (a[i] < b[i]) return -1;
    }
    return 0;
}

//高精度 a * 低精度 x = 高精度 c
void mul(int a[], int x, int c[]) {
    int t = 0;
    c[0] = a[0];

    for (int i = 1; i <= c[0]; i ++ ) {
        t += a[i] * x;
        c[i] = t % 10;
        t /= 10;
    }

    while (t) c[++c[0]] = t % 10, t /= 10;
    while (!c[c[0]] && c[0] > 1) c[0] -- ;
}

//高精度 a / 低精度 x = 高精度 c，余数为 r
void div(int a[], int x, int c[], int &r) {
    r = 0;
    c[0] = a[0];

    for (int i = c[0]; i >= 1; i -- ) {
        r = r * 10 + a[i];
        c[i] = r / x;
        r %= x;
    }

    while (!c[c[0]] && c[0] > 1)  c[0] -- ;
}

//高精度 a / 高精度 b = 高精度 c，余数为 r
void div2(int a[], int b[], int c[]) {
    c[0] = max(a[0] - b[0] + 1, 1);

    while(b[0] < a[0]) mul(b, 10, b);

    for (int i = c[0]; i >= 1; i -- ) {
        while (cmp(a,b) != -1) {
            c[i] ++ ;
            sub(a, b, a);
        }
        int r;
        div(b, 10, b, r);
    }

    while (!c[c[0]] && c[0] > 1)  c[0] -- ;
}

int main() {
    init(a);
    init(b);

    div2(a, b, c);

    dy(c);
    dy(a);

    return 0;
}

1309 palindrome

#include <iostream>
using namespace std;

const int N = 1e5;
int a[N], c[N], n, cnt = 0;
string m;

void add() {
    int t = 0;

    for (int i = 1, j = a[0]; i <= a[0]; i ++, j -- ) {
        t = t + a[i] + a[j];
        c[i] = t % n;
        t /= n;
    }
    if (t) c[++a[0]] = t;

    for (int i = 1; i <= a[0]; i ++ ) {
        a[i] = c[i];
    }
}

bool isPalindromic() {
    for (int i = 1, j = a[0]; i < j; i ++, j -- ) {
        if (a[i] != a[j]) return false;
    }
    return true;
}

int main() {
    cin >> n >> m;

    a[0] = m.size();
    for (int i = m.size(); i >= 1; i -- ) {
        char ch = m[a[0] - i];
        if (ch >= 'A') a[i] = ch - 'A' + 10;
        else a[i] = ch - '0';
    }

    bool res = isPalindromic();
    while (!res && cnt <= 30) {
        add();
        cnt ++ ;
        res = isPalindromic();
    }

    if (cnt <= 30) cout << cnt;
    else cout << "Impossible";

    return 0;
}

1168 Large integer addition

#include <iostream>
using namespace std;

const int N = 1e5 + 10;
int a[N], b[N], c[N], x;

void init(int a[]) {
    string s;
    cin >> s;

    a[0] = s.size();
    for (int i = 1; i <= a[0]; i ++ ) {
        a[i] = s[a[0]-i] - '0';
    }
}

void dy(int a[]) {
    for (int i = a[0]; i >= 1; i --) {
        cout << a[i];
    }

    cout << endl;
}

//高精度 a + 高精度 b = 高精度 c
void add(int a[], int b[], int c[]) {
    int t = 0;
    c[0] = max(a[0], b[0]);

    for (int i = 1; i <= c[0]; i ++ ) {
        t += a[i] + b[i];
        c[i] = t % 10;
        t /= 10;
    }

    if (t) c[++c[0]] = t;
    while (!c[c[0]] && c[0] > 1) c[0] -- ;
}

int main() {
    init(a);
    init(b);

    add(a, b, c);

    dy(c);

    return 0;
}

1169 Large integer subtraction

#include <iostream>
using namespace std;

const int N = 1e5 + 10;
int a[N], b[N], c[N], x;

void init(int a[]) {
    string s;
    cin >> s;

    a[0] = s.size();
    for (int i = 1; i <= a[0]; i ++ ) {
        a[i] = s[a[0]-i] - '0';
    }
}

void dy(int a[]) {
    for (int i = a[0]; i >= 1; i --) {
        cout << a[i];
    }

    cout << endl;
}

//高精度 a - 高精度 b = 高精度 c
void sub(int a[], int b[], int c[]) {
    int t = 0;
    c[0] = a[0];

    for (int i = 1; i <= c[0]; i ++ ) {
        t = a[i] - b[i] - t;
        c[i] = (t + 10) % 10;
        t = t < 0 ? 1 : 0;
    }
    while (!c[c[0]] && c[0] > 1) c[0] -- ;
}

int main() {
    init(a);
    init(b);

    sub(a, b, c);

    dy(c);

    return 0;
}

1170 Calculate the n-th power of 2

#include <iostream>
using namespace std;

const int N = 1e5 + 10;
int a[N], b[N], c[N], x;

void dy(int a[]) {
    for (int i = a[0]; i >= 1; i --) {
        cout << a[i];
    }

    cout << endl;
}

//高精度 a * 低精度 x = 高精度 c
void mul(int a[], int x, int c[]) {
    int t = 0;
    c[0] = a[0];

    for (int i = 1; i <= c[0]; i ++ ) {
        t += a[i] * x;
        c[i] = t % 10;
        t /= 10;
    }

    while (t) c[++c[0]] = t % 10, t /= 10;
    while (!c[c[0]] && c[0] > 1) c[0] -- ;
}

int main() {
    a[0] = 1, a[1] = 1;

    int n;
    cin >> n;

    while (n--)    {
        mul(a, 2, a);
    }

    dy(a);

    return 0;
}

Factors of 1171 large integers

#include <iostream>
using namespace std;

const int N = 1e5 + 10;
int a[N], b[N], c[N], x;

void init(int a[]) {
    string s;
    cin >> s;

    a[0] = s.size();
    for (int i = 1; i <= a[0]; i ++ ) {
        a[i] = s[a[0]-i] - '0';
    }
}

//高精度 a / 低精度 x = 高精度 c，余数为 r
void div(int a[], int x, int c[], int &r) {
    r = 0;
    c[0] = a[0];

    for (int i = c[0]; i >= 1; i -- ) {
        r = r * 10 + a[i];
        c[i] = r / x;
        r %= x;
    }

    while (!c[c[0]] && c[0] > 1)  c[0] -- ;
}

int main() {
    init(a);

    int r;
    bool flag = true;
    for (int i = 2; i <= 9; i ++ ) {
        div(a, i, c, r);
        if (!r) {
            cout << i << ' ';
            flag = false;
        }
    }

    if (flag)  cout << "none";

    return 0;
}

1172 Find the factorial of n within 10000

#include <iostream>
using namespace std;

const int N = 1e5 + 10;
int a[N], b[N], c[N], x;

void dy(int a[]) {
    for (int i = a[0]; i >= 1; i --) {
        cout << a[i];
    }

    cout << endl;
}

//高精度 a * 低精度 x = 高精度 c
void mul(int a[], int x, int c[]) {
    int t = 0;
    c[0] = a[0];

    for (int i = 1; i <= c[0]; i ++ ) {
        t += a[i] * x;
        c[i] = t % 10;
        t /= 10;
    }

    while (t) c[++c[0]] = t % 10, t /= 10;
    while (!c[c[0]] && c[0] > 1) c[0] -- ;
}

int main() {
    a[0] = 1, a[1] = 1;

    int n;
    cin >> n;

    for (int i = 1; i <= n; i ++ ) {
        mul(a, i, a);
    }

    dy(a);

    return 0;
}

1173 Factorial Sum

#include <iostream>
using namespace std;

const int N = 1e5 + 10;
int a[N], b[N], c[N], x;

void dy(int a[]) {
    for (int i = a[0]; i >= 1; i --) {
        cout << a[i];
    }

    cout << endl;
}

//高精度 a + 高精度 b = 高精度 c
void add(int a[], int b[], int c[]) {
    int t = 0;
    c[0] = max(a[0], b[0]);

    for (int i = 1; i <= c[0]; i ++ ) {
        t += a[i] + b[i];
        c[i] = t % 10;
        t /= 10;
    }

    if (t) c[++c[0]] = t;
    while (!c[c[0]] && c[0] > 1) c[0] -- ;
}

//高精度 a * 低精度 x = 高精度 c
void mul(int a[], int x, int c[]) {
    int t = 0;
    c[0] = a[0];

    for (int i = 1; i <= c[0]; i ++ ) {
        t += a[i] * x;
        c[i] = t % 10;
        t /= 10;
    }

    while (t) c[++c[0]] = t % 10, t /= 10;
    while (!c[c[0]] && c[0] > 1) c[0] -- ;
}

int main() {
    a[0] = 1, a[1] = 1;

    int n;
    cin >> n;

    for (int i = 1; i <= n; i ++ ) {
        mul(a, i, a);
        add(c, a, c);
    }

    dy(c);

    return 0;
}

1175 divided by 13

#include <iostream>
using namespace std;

const int N = 1e5 + 10;
int a[N], b[N], c[N], x;

void init(int a[]) {
    string s;
    cin >> s;

    a[0] = s.size();
    for (int i = 1; i <= a[0]; i ++ ) {
        a[i] = s[a[0]-i] - '0';
    }
}

void dy(int a[]) {
    for (int i = a[0]; i >= 1; i --) {
        cout << a[i];
    }

    cout << endl;
}

//高精度 a / 低精度 x = 高精度 c，余数为 r
void div(int a[], int x, int c[], int &r) {
    r = 0;
    c[0] = a[0];

    for (int i = c[0]; i >= 1; i -- ) {
        r = r * 10 + a[i];
        c[i] = r / x;
        r %= x;
    }

    while (!c[c[0]] && c[0] > 1)  c[0] -- ;
}

int main() {
    init(a);

    int r;
    div(a, 13, c, r);

    dy(c);
    cout << r;

    return 0;
}

If your child is in the fourth grade or above, is interested in computer programming and has spare capacity in cultural lessons, please contact customer service (WeChat ID: xiaolan7321) to participate in informatics learning. We are professional informatics competition coaches. We use online small class teaching methods. The goal is to help primary and middle school students who love programming to achieve excellent results in informatics competitions at home and abroad.

Teaching features:

• Online small class teaching, lay a good code foundation. Avoid the problem of “cannot keep up” or “not enough to eat” in large classes.

• Rich teaching experience, familiar with students' knowledge structure and learning ability, and arrange the schedule reasonably.

• Practice with competitions, and continuously improve students' abilities through examinations and competitions.