145 条题解
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极简版：
#include <iostream> using namespace std; int main(){int a,b;cin>>a>>b;cout<<a+b;return 0;} -
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SunnyLi @ 2023-10-2 13:53:11
#include<bits/stdc++.h> //引入头文件 using namespace std; //设置namespace int main(){ //主函数 int a,b; //定义变量 cin >>a >> b; //输入 cout << a+b; //输出 return 0; //结束代码 } -
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A+B Problem
题解+思路
题解
#include <iostream> using namespace std; struct Node { int data; Node *prev; Node *next; Node(int val) : data(val), prev(nullptr), next(nullptr) {} }; Node *createList(int num) { Node *head = nullptr; Node *tail = nullptr; while (num > 0) { int digit = num % 10; Node *newNode = new Node(digit); if (head == nullptr) { head = newNode; tail = newNode; } else { newNode->next = head; head->prev = newNode; head = newNode; } num /= 10; } return head; } Node *addTwoNumbers(Node *num1, Node *num2) { Node *result = nullptr; Node *current = nullptr; int carry = 0; while (num1 != nullptr || num2 != nullptr || carry != 0) { int sum = carry; if (num1 != nullptr) { sum += num1->data; num1 = num1->next; } if (num2 != nullptr) { sum += num2->data; num2 = num2->next; } carry = sum / 10; sum %= 10; Node *newNode = new Node(sum); if (result == nullptr) { result = newNode; current = newNode; } else { current->prev = newNode; newNode->next = current; current = newNode; } } return result; } void printList(Node *head) { if (head == nullptr) { cout << "Empty list" << endl; return; } while (head != nullptr) { cout << head->data; head = head->next; } cout << endl; } void deleteList(Node *head) { while (head != nullptr) { Node *temp = head; head = head->next; delete temp; } } int main() { int num1 = 12345; int num2 = 6789; Node *list1 = createList(num1); Node *list2 = createList(num2); cout << "Number 1: "; printList(list1); cout << "Number 2: "; printList(list2); Node *sumList = addTwoNumbers(list1, list2); cout << "Sum: "; printList(sumList); deleteList(list1); deleteList(list2); deleteList(sumList); return 0; }思路
- 首先，定义一个双向链表的节点结构体，包含一个整数值和两个指针，分别指向前一个节点和后一个节点。
- 创建两个双向链表，分别表示要相加的两个数字。可以通过遍历输入的整数，从个位开始，逐个创建节点并将其插入链表中。
- 定义一个变量来表示进位，初始值为0。
- 从链表的最后一个节点开始，逐位相加，并将结果存储在新的链表中。具体步骤如下： - 从两个链表的最后一个节点开始，分别取出对应的值，并加上进位。 - 将相加的结果对10取模得到当前位的值，同时更新进位为相加结果除以10的商。 - 创建一个新的节点，将当前位的值存储在该节点中，并将该节点插入到结果链表的头部。 - 分别将两个链表的指针指向上一个节点，继续下一位的相加操作。 - 如果其中一个链表已经遍历完，但另一个链表还有剩余位数，则将剩余位数与进位相加，并将结果插入到结果链表中
- 最后，得到的结果链表即为相加后的结果。
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Michael_114 LV 7 @ 2024-8-19 16:20:15
本蒟蒻啥都不会，只能贡献一份小小的码风优良的代码```language
#include <iostream> using namespace std; const int MAX = 1000; void change(const string &a1, int a[], int len) { for (int i = 0; i < len; i++) { a[i] = a1[len - i - 1] - '0'; } } int main() { string a1, b1; cin >> a1 >> b1; int lena = a1.size(), lenb = b1.size(); int a[MAX] = {}, b[MAX] = {}, c[MAX] = {}, cf = 0; change(a1, a, lena); change(b1, b, lenb); int lenc = lena; if (lenb > lena) { lenc = lenb; } for (int i = 0; i <= lenc; i++) { c[i] = a[i] + b[i] + cf; if (c[i] >= 10) { cf = 1; c[i] -= 10; } else { cf = 0; } } int i = lenc; if (c[i] == 0) { i--; } for (i; i >= 0; i--) { cout << c[i]; } return 0; } -
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#include<bits/stdc++.h> //头文件 using namespace std;//命名空间 int main(){//主函数,程序从这里开始 //signed main也行 long long a,b; //用long long类型的变量进行存储因为数值范围是10^6，int存不下 cin>>a>>b; /* 这里可以替换为： scanf("%lld%lld",&a,&b); */ long long ans=a+b; cout<<ans<<endl; /* 这里可以替换为： printf("%lld",ans); */ return 0;//一定要写,不然RE } -
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终于可以写一份A+B这么难的题的题解了。 咦？竟然没有人写LCT的题解？ Link-Cut Tree 会很伤心的！ ORZ为了不让LCT伤心于是我来一份LCT的A+B题解吧！ 送上代码： #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; }本蒟蒻第一次写题解，请原谅
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Minelimer @ 2024-7-25 19:06:25
又一个抄洛谷的🤣👉
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haohao0920 @ 2024-7-1 20:24:17
#include<bits/stdc++.h> using namespace std; int main() { int a, b; cin>>a>>b; cout<<a+b; return 0; } -
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#include <bits/stdc++.h> using namespace std; const int N = 1e4 + 15, M = 2e5 + 10; vector<int> edge[N]; int h[N], e[M], ne[M], idx, p[N]; int in[N]; int n, m; void add(int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ; in[b]++; } int dfn[N], low[N], tot = 0; stack<int> stk; bool st[N]; int sz[N], scc[N], color; void tarjan(int u) { dfn[u] = low[u] = ++tot; stk.push(u); st[u] = true; for (int v : edge[u]) { if (!dfn[v]) tarjan(v), low[u] = min(low[u], low[v]); else if (st[v]) low[u] = min(low[u], dfn[v]); } if (dfn[u] == low[u]) { ++color; while (stk.top() != u) { scc[stk.top()] = color; st[stk.top()] = false; stk.pop(); } scc[stk.top()] = color; st[stk.top()] = false; stk.pop(); } } queue<int> q; int dp[N]; void topsort() { for (int i = 1; i <= color; i++) { dp[i] = sz[i]; if (in[i] == 0) q.push(i); } while (q.size()) { int u = q.front(); q.pop(); for (int i = h[u]; ~i; i = ne[i]) { int v = e[i]; dp[v] = max(dp[v], dp[u] + sz[v]); in[v]--; if (in[v] == 0) q.push(v); } } int ans = 0; for (int i = 1; i <= color; i++) ans = max(ans, dp[i]); printf("%d\n", ans); } int main() { memset(h, -1, sizeof h); n = 2, m = 1; for (int i = 1; i <= n; i++) scanf("%d", &p[i]); while (m--) { int a = 1, b = 2; edge[a].push_back(b); } for (int i = 1; i <= n; i++) if (!dfn[i]) tarjan(i); for (int i = 1; i <= n; i++) { sz[scc[i]] += p[i]; for (int j : edge[i]) if (scc[i] != scc[j]) add(scc[i], scc[j]); } topsort(); return 0; }转自AcWing@Conan15
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孙浩匀
@ 2024-5-19 21:03:56
A+B作为一道基础题，思维也很简单 看代码：
#include <bits/stdc++.h> #define endl '\n' #define int long long using namespace std; signed main() { int a, b; cin >> a >> b; cout << a + b; return 0; }
张汤明甫
LV 6
@ 2023-10-3 19:28:15
