1. A + B Problem（题解）

第一种方法：
#include<iostream>
#include<cstdio>
using namespace std;
int main(){
    int a,b;
    scanf("%d",&a,&b);
    printf("%d",a+b);
}

第二种方法：
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
using namespace std;
struct node {
    int data, rev, sum;
    node *son[2], *pre;
    bool judge();
    bool isroot();
    void pushdown();
    void update();
    void setson(node *child, int lr);
} lct[233];
int top, a, b;
node *getnew(int x) {
    node *now = lct + ++top;
    now->data = x;
    now->pre = now->son[1] = now->son[0] = lct;
    now->sum = 0;
    now->rev = 0;
    return now;
}
bool node::judge() { return pre->son[1] == this; }
bool node::isroot() {
    if (pre == lct)
        return true;
    return !(pre->son[1] == this || pre->son[0] == this);
}
void node::pushdown() {
    if (this == lct || !rev)
        return;
    swap(son[0], son[1]);
    son[0]->rev ^= 1;
    son[1]->rev ^= 1;
    rev = 0;
}
void node::update() { sum = son[1]->sum + son[0]->sum + data; }
void node::setson(node *child, int lr) {
    this->pushdown();
    child->pre = this;
    son[lr] = child;
    this->update();
}
void rotate(node *now) {
    node *father = now->pre, *grandfa = father->pre;
    if (!father->isroot())
        grandfa->pushdown();
    father->pushdown();
    now->pushdown();
    int lr = now->judge();
    father->setson(now->son[lr ^ 1], lr);
    if (father->isroot())
        now->pre = grandfa;
    else
        grandfa->setson(now, father->judge());
    now->setson(father, lr ^ 1);
    father->update();
    now->update();
    if (grandfa != lct)
        grandfa->update();
}
void splay(node *now) {
    if (now->isroot())
        return;
    for (; !now->isroot(); rotate(now))
        if (!now->pre->isroot())
            now->judge() == now->pre->judge() ? rotate(now->pre) : rotate(now);
}
node *access(node *now) {
    node *last = lct;
    for (; now != lct; last = now, now = now->pre) {
        splay(now);
        now->setson(last, 1);
    }
    return last;
}
void changeroot(node *now) {
    access(now)->rev ^= 1;
    splay(now);
}
void connect(node *x, node *y) {
    changeroot(x);
    x->pre = y;
    access(x);
}
void cut(node *x, node *y) {
    changeroot(x);
    access(y);
    splay(x);
    x->pushdown();
    x->son[1] = y->pre = lct;
    x->update();
}
int query(node *x, node *y) {
    changeroot(x);
    node *now = access(y);
    return now->sum;
}
int main() {
    scanf("%d%d", &a, &b);
    node *A = getnew(a);
    node *B = getnew(b);
    //连边 Link
    connect(A, B);
    //断边 Cut
    cut(A, B);
    //再连边orz Link again
    connect(A, B);
    printf("%d\n", query(A, B));
    return 0;
}

第三个方法（结构体高精）：
#include <bits/stdc++.h>
using namespace std;
struct bignumber {
    int len, a[10005];
    bignumber() {
        memset(a, 0, sizeof(a));
        len = 1;
    }
    void read() {
        string s;
        cin >> s;
        len = s.size();
        for (int i = 0; i <= len - 1; i++) a[len - i] = s[i] - '0';
    }
    void print() {
        for (int i = len; i >= 1; i--) cout << a[i];
    }
    friend bignumber operator+(bignumber x, bignumber y) {
        bignumber z;
        z.len = max(x.len, y.len);
        for (int i = 1; i <= z.len; i++) {
            z.a[i] = x.a[i] + y.a[i];
        }
        return fix(z);
    }
    friend bignumber fix(bignumber x) {
        for (int i = 1; i < x.len; i++) {
            x.a[i + 1] += x.a[i] / 10;
            x.a[i] %= 10;
        }

        if (x.a[x.len] > 9) {
            x.a[x.len + 1] += x.a[x.len] / 10;
            x.a[x.len] %= 10;
            x.len++;
        }
        while (x.len > 1 && x.a[x.len] == 0) x.len--;
        return x;
    }
};
int main() {
    bignumber x, y, z;
    x.read();
    y.read();
    z = x + y;
    z.print();
    return 0;
}