C++ A+B problem答案

CPP 刷题王 2022-11-22 12:27:42 2022-11-22 12:34:14 6

这是你以为的:

#include <iostream>
using namespace std;
int main()
{
 int a, b;
 cin >> a >> b;
 cout << a + b<<endl;
 return 0;
}

这是大佬们的: 1号

#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);
	connect(A, B);
	cut(A, B);
	connect(A, B);
	printf("%d\n", query(A, B));
	return 0;
}

2号:

#include<iostream>
#include<cstring>
using namespace std;
int lowbit(int a)
{
    return a&(-a);
}
int main()
{
    int n=2,m=1;
    int ans[m+1];
    int a[n+1],c[n+1],s[n+1];
    int o=0;
    memset(c,0,sizeof(c));
    s[0]=0;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        s[i]=s[i-1]+a[i];
        c[i]=s[i]-s[i-lowbit(i)];//树状数组创建前缀和优化
    }
    for(int i=1;i<=m;i++)
    {
        int q=2;
        //if(q==1)
        //{(没有更改操作)
        //    int x,y;
        //    cin>>x>>y;
        //    int j=x;
        //    while(j<=n)
        //    {
        //        c[j]+=y;
        //        j+=lowbit(j);
        //    }
        //}
        //else
        {
            int x=1,y=2;//求a[1]+a[2]的和
            int s1=0,s2=0,p=x-1;
            while(p>0)
            {
                s1+=c[p];
                p-=lowbit(p);//树状数组求和操作,用两个前缀和相减得到区间和
            }
            p=y;
            while(p>0)
            {
                s2+=c[p];
                p-=lowbit(p);
            }    
            o++;
            ans[o]=s2-s1;//存储答案
        }
    }
    for(int i=1;i<=o;i++)
        cout<<ans[i]<<endl;//输出
    return 0;
}

3号:

#include <bits/stdc++.h>
#define ll long long
#define N 100000
using namespace std;
int sz[N], rev[N], tag[N], sum[N], ch[N][2], fa[N], val[N];
int n, m, rt, x;
void push_up(int x){
    sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1;
    sum[x] = sum[ch[x][1]] + sum[ch[x][0]] + val[x];
}
void push_down(int x){
    if(rev[x]){
        swap(ch[x][0], ch[x][1]);
        if(ch[x][1]) rev[ch[x][1]] ^= 1;
        if(ch[x][0]) rev[ch[x][0]] ^= 1;
        rev[x] = 0;
    }
    if(tag[x]){
        if(ch[x][1]) tag[ch[x][1]] += tag[x], sum[ch[x][1]] += tag[x];
        if(ch[x][0]) tag[ch[x][0]] += tag[x], sum[ch[x][0]] += tag[x];
        tag[x] = 0;
    }
}
void rotate(int x, int &k){
    int y = fa[x], z = fa[fa[x]];
    int kind = ch[y][1] == x;
    if(y == k) k = x;
    else ch[z][ch[z][1]==y] = x;
    fa[x] = z; fa[y] = x; fa[ch[x][!kind]] = y;
    ch[y][kind] = ch[x][!kind]; ch[x][!kind] = y;
    push_up(y); push_up(x);
}
void splay(int x, int &k){
    while(x != k){
        int y = fa[x], z = fa[fa[x]];
        if(y != k) if(ch[y][1] == x ^ ch[z][1] == y) rotate(x, k);
        else rotate(y, k);
        rotate(x, k);
    }
}
int kth(int x, int k){
    push_down(x);
    int r = sz[ch[x][0]]+1;
    if(k == r) return x;
    if(k < r) return kth(ch[x][0], k);
    else return kth(ch[x][1], k-r);
}
void split(int l, int r){
    int x = kth(rt, l), y = kth(rt, r+2);
    splay(x, rt); splay(y, ch[rt][1]);
}
void rever(int l, int r){
    split(l, r);
    rev[ch[ch[rt][1]][0]] ^= 1;
}
void add(int l, int r, int v){
    split(l, r);
    tag[ch[ch[rt][1]][0]] += v;
    val[ch[ch[rt][1]][0]] += v;
    push_up(ch[ch[rt][1]][0]);
}
int build(int l, int r, int f){
    if(l > r) return 0;
    if(l == r){
        fa[l] = f;
        sz[l] = 1;
        return l;
    }
    int mid = l + r >> 1;
    ch[mid][0] = build(l, mid-1, mid);
    ch[mid][1] = build(mid+1, r, mid);
    fa[mid] = f;
    push_up(mid);
    return mid;
}
int asksum(int l, int r){
    split(l, r);
    return sum[ch[ch[rt][1]][0]];
}
int main(){
    //总共两个数
    n = 2;
    rt = build(1, n+2, 0);//建树
    for(int i = 1; i <= n; i++){
        scanf("%d", &x);
        add(i, i, x);//区间加
    }
    rever(1, n);//区间翻转
    printf("%d\n", asksum(1, n));//区间求和
    return 0;
}

4号:

#include <iostream> 
#include <cmath>
using namespace std;
int fu=1,f=1,a,b,c=0;
int main()
{
    cin>>a>>b;
    if(a<0&&b>0)fu=2;
    if(a>0&&b<0)fu=3;
    if(a<0&&b<0)f=-1;
    if(a==0){cout<<b;return 0;}
    if(b==0){cout<<a;return 0;} 
    a=abs(a);
    b=abs(b);
    if(a>b&&fu==3)f=1;
    if(b>a&&fu==3)f=-1;
    if(b>a&&fu==2)f=1;
    if(b<a&&fu==2)f=-1;
    if(fu==1)c=a+b;
    if(fu>1)c=max(a,b)-min(a,b);
    c*=f;
    cout<<c;
    return 0;
}

5号:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <climits>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
#include <ctime>
#include <string>
#include <cstring>
using namespace std;
const int N=405;
struct Edge {
    int v,w;
};
vector<Edge> edge[N*N];
int n;
int dis[N*N];
bool vis[N*N];
struct cmp {
    bool operator()(int a,int b) {
        return dis[a]>dis[b];
    }
};
int Dijkstra(int start,int end)
{
    priority_queue<int,vector<int>,cmp> dijQue;
    memset(dis,-1,sizeof(dis));
    memset(vis,0,sizeof(vis));
    dijQue.push(start);
    dis[start]=0;
    while(!dijQue.empty()) {
        int u=dijQue.top();
        dijQue.pop();
        vis[u]=0;
        if(u==end)
            break;
        for(int i=0; i<edge[u].size(); i++) {
            int v=edge[u][i].v;
            if(dis[v]==-1 || dis[v]>dis[u]+edge[u][i].w) {
                dis[v]=dis[u]+edge[u][i].w;
                if(!vis[v]) {
                    vis[v]=true;
                    dijQue.push(v);
                }
            }
        }
    }
    return dis[end];
}
int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    Edge Qpush;
Qpush<span class="hl-punctuation hl-accessor hl-dot hl-c++">.</span><span class="hl-variable hl-other hl-readwrite hl-member hl-c++">v</span><span class="hl-keyword hl-operator hl-assignment hl-c">=</span><span class="hl-constant hl-numeric hl-c++">1</span><span class="hl-punctuation hl-terminator hl-c++">;</span>
Qpush<span class="hl-punctuation hl-accessor hl-dot hl-c++">.</span><span class="hl-variable hl-other hl-readwrite hl-member hl-c++">w</span><span class="hl-keyword hl-operator hl-assignment hl-c">=</span>a<span class="hl-punctuation hl-terminator hl-c++">;</span>
edge<span class="hl-meta hl-brackets hl-c++"><span class="hl-punctuation hl-section hl-brackets hl-begin hl-c++">[</span><span class="hl-constant hl-numeric hl-c++">0</span><span class="hl-punctuation hl-section hl-brackets hl-end hl-c++">]</span></span><span class="hl-punctuation hl-accessor hl-dot hl-c++">.</span><span class="hl-meta hl-method-call hl-c++"><span class="hl-variable hl-function hl-member hl-c++">push_back</span><span class="hl-meta hl-group hl-c++"><span class="hl-punctuation hl-section hl-group hl-begin hl-c++">(</span></span></span><span class="hl-meta hl-method-call hl-c++"><span class="hl-meta hl-group hl-c++">Qpush</span></span><span class="hl-meta hl-method-call hl-c++"><span class="hl-meta hl-group hl-c++"><span class="hl-punctuation hl-section hl-group hl-end hl-c++">)</span></span></span><span class="hl-punctuation hl-terminator hl-c++">;</span>

Qpush<span class="hl-punctuation hl-accessor hl-dot hl-c++">.</span><span class="hl-variable hl-other hl-readwrite hl-member hl-c++">v</span><span class="hl-keyword hl-operator hl-assignment hl-c">=</span><span class="hl-constant hl-numeric hl-c++">2</span><span class="hl-punctuation hl-terminator hl-c++">;</span>
Qpush<span class="hl-punctuation hl-accessor hl-dot hl-c++">.</span><span class="hl-variable hl-other hl-readwrite hl-member hl-c++">w</span><span class="hl-keyword hl-operator hl-assignment hl-c">=</span>b<span class="hl-punctuation hl-terminator hl-c++">;</span>
edge<span class="hl-meta hl-brackets hl-c++"><span class="hl-punctuation hl-section hl-brackets hl-begin hl-c++">[</span><span class="hl-constant hl-numeric hl-c++">1</span><span class="hl-punctuation hl-section hl-brackets hl-end hl-c++">]</span></span><span class="hl-punctuation hl-accessor hl-dot hl-c++">.</span><span class="hl-meta hl-method-call hl-c++"><span class="hl-variable hl-function hl-member hl-c++">push_back</span><span class="hl-meta hl-group hl-c++"><span class="hl-punctuation hl-section hl-group hl-begin hl-c++">(</span></span></span><span class="hl-meta hl-method-call hl-c++"><span class="hl-meta hl-group hl-c++">Qpush</span></span><span class="hl-meta hl-method-call hl-c++"><span class="hl-meta hl-group hl-c++"><span class="hl-punctuation hl-section hl-group hl-end hl-c++">)</span></span></span><span class="hl-punctuation hl-terminator hl-c++">;</span>

<span class="hl-meta hl-function-call hl-c++"><span class="hl-support hl-function hl-C99 hl-c">printf</span><span class="hl-meta hl-group hl-c++"><span class="hl-punctuation hl-section hl-group hl-begin hl-c++">(</span></span></span><span class="hl-meta hl-function-call hl-c++"><span class="hl-meta hl-group hl-c++"><span class="hl-string hl-quoted hl-double hl-c"><span class="hl-punctuation hl-definition hl-string hl-begin hl-c">"</span><span class="hl-constant hl-other hl-placeholder hl-c">%d</span><span class="hl-punctuation hl-definition hl-string hl-end hl-c">"</span></span><span class="hl-punctuation hl-separator hl-c++">,</span><span class="hl-meta hl-function-call hl-c++"><span class="hl-variable hl-function hl-c++">Dijkstra</span><span class="hl-meta hl-group hl-c++"><span class="hl-punctuation hl-section hl-group hl-begin hl-c++">(</span></span></span><span class="hl-meta hl-function-call hl-c++"><span class="hl-meta hl-group hl-c++"><span class="hl-constant hl-numeric hl-c++">0</span><span class="hl-punctuation hl-separator hl-c++">,</span><span class="hl-constant hl-numeric hl-c++">2</span></span></span><span class="hl-meta hl-function-call hl-c++"><span class="hl-meta hl-group hl-c++"><span class="hl-punctuation hl-section hl-group hl-end hl-c++">)</span></span></span></span></span><span class="hl-meta hl-function-call hl-c++"><span class="hl-meta hl-group hl-c++"><span class="hl-punctuation hl-section hl-group hl-end hl-c++">)</span></span></span><span class="hl-punctuation hl-terminator hl-c++">;</span>
<span class="hl-keyword hl-control hl-c++">return</span> <span class="hl-constant hl-numeric hl-c++">0</span><span class="hl-punctuation hl-terminator hl-c++">;</span>

}

6号:

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
struct node{
    int str[26];
    int sum;
}s[1000];
char str1[100];
int t=0,tot=0,ss=0;
bool f1;
void built()
{
    t=0;
    for(int i=0;i<strlen(str1);i++)
    {
         if(str1[i]=='-'){
             f1=true;continue;
         }
         if(!s[t].str[str1[i]-'0'])
         s[t].str[str1[i]-'0']=++tot;
         t=s[t].str[str1[i]-'0'];
         s[t].sum=str1[i]-'0';
    }
}
int query()
{
   int t=0;int s1=0;
   for(int i=0;i<strlen(str1);i++)
   {
           if(str1[i]=='-') continue;
           if(!s[t].str[str1[i]-'0']) return s1;
           t=s[t].str[str1[i]-'0'];
           s1=s1*10+s[t].sum;
   }
   return s1;
}
int main()
{    
  for(int i=1;i<=2;i++)
  {
      f1=false;
      scanf("%s",str1);
    built();
    if(f1)
      ss-=query();
      else ss+=query();
  }
  printf("%d",ss);
  return 0;    
}

7号:

#include<cstdio>
#define reg register
inline int s()
{
    reg char ch=getchar();
    reg int re=0;
    reg bool fl=1;
    if(ch=='-')
    {
        fl=0;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        re=re*10+ch-'0';
        ch=getchar();
    }
    return fl?re:-re;
}
inline bool w(reg int r)
{
    if(r>9)
        w(r/10);
    putchar(r%10+'0');
    return 1;
}
int main()
{
    reg int a=s(),b=s();
    if(a+b>=0)
        return !w(a+b);
    putchar('-');
    return !w(-a-b);
}

8号

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
int main()
{
    int a,b,s=0,s1=0,i=0,na=0,nb=0;
    cin>>a>>b;
    if(a<=0) na=1,a*=-1;
    while(a!=0)
    {
        if(a%2!=0)
        s+=pow(2,a%2*i);
        a/=2;
        i++;
    }
    i=0;
    if(na==1) s*=-1;
    if(b<=0) nb=1,b*=-1;
    while(b!=0)
    {
        if(b%2!=0)
        s1+=pow(2,b%2*i);
        b/=2;
        i++;
    }
    if(nb==1) s1*=-1;
    cout<<s+s1;;
    return 0;
}

9号:

#include <cstdio>
int read()
{
    int f=1,x=0;char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}
    return f*x;
}
int main()
{
    printf("%d",read()+read());
    return 0;
}

10号:

#include <iostream>
using namespace std;
int plus(int a,int b)//这个是加法运算函数
{
    if(b==0)//如果b(进位)是0(没有进位了),返回a的值
        return a;
    else
    {
        int xor,carry;
        xor=a^b;//xor是a和b不进位加法的值
        carry=(a&b)<<1;//carry是a和b进位的值(只有两个都是1才会产生进位,所以是与运算。左移一位是因为二进制加法和十进制加法竖式一样进位要加在左面一位里)
        return plus(xor,carry);//把不进位加法和进位的值的和就是结果
    }
}
int main()
{
    int a,b;
    cin >> a >> b;
    cout << plus(a,b) << endl;
    return 0;
}

11号:

#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<iostream>
using namespace std;
struct node{
    int val,l,r;
};
node t[5];
int a[5],f[5];
int n,m;
void init(){
    for(int i=1;i<=2;i++){
        scanf("%d",&a[i]);
    }
}
void build(int l,int r,int node){//这是棵树
    t[node].l=l;t[node].r=r;t[node].val=0;
    if(l==r){
        f[l]=node;
        t[node].val=a[l];
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,node*2);
    build(mid+1,r,node*2+1);
    t[node].val=t[node*2].val+t[node*2+1].val;
}
void update(int node){
    if(node==1)return;
    int fa=node>>1;
    t[fa].val=t[fa*2].val+t[fa*2+1].val;
    update(fa);
}
int find(int l,int r,int node){
    if(t[node].l==l&&t[node].r==r){
        return t[node].val;
    }
    int sum=0;
    int lc=node*2;int rc=lc+1;
    if(t[lc].r>=l){
        if(t[lc].r>=r){
            sum+=find(l,r,lc);
        }
        else{
            sum+=find(l,t[lc].r,lc);
        }
    }
    if(t[rc].l<=r){
        if(t[rc].l<=l){
            sum+=find(l,r,rc);
        }
        else{
            sum+=find(t[rc].l,r,rc);
        }
    }
    return sum;
}
int main(){
    init();
    build(1,2,1);
    printf("%d",find(1,2,1));
}

12号:

#include <cstdio>
int m, n;
int main()
{
    scanf("%d%d", &m, &n);
    int u = m & n;
    int v = m ^ n;
    while (u) {
        int s = v;
        int t = u << 1;
        u = s & t;
        v = s ^ t;
    }
    printf("%d\n", v);
}

13号

#include<cstdio>                                                  //头文件
#define NI 2                                                          
//从来不喜欢算log所以一般用常数 不知道算不算坏习惯 因为3个节点 所以log3(当然以2为底)上取整得2
struct edge
{
    int to,next,data;                                              //分别表示边的终点,下一条边的编号和边的权值
}e[30];                                                                     //邻接表,点少边少开30是为了浪啊
int v[10],d[10],lca[10][NI+1],f[10][NI+1],tot=0;      //数组开到10依然为了浪
//数组还解释嘛,v表示第一条边在邻接表中的编号,d是深度,lca[x][i]表示x向上跳2^i的节点,f[x][i]表示x向上跳2^i的距离和
void build(int x,int y,int z)                                      //建边
{
    e[++tot].to=y; e[tot].data=z; e[tot].next=v[x]; v[x]=tot;
    e[++tot].to=x; e[tot].data=z; e[tot].next=v[y]; v[y]=tot;
}
void dfs(int x)                                                        //递归建树
{
    for(int i=1;i<=NI;i++)                                   //懒,所以常数懒得优化
        f[x][i]=f[x][i-1]+f[lca[x][i-1]][i-1],
        lca[x][i]=lca[lca[x][i-1]][i-1];                   //建树的同时进行预处理
    for(int i=v[x];i;i=e[i].next)                              //遍历每个连接的点
    {
        int y=e[i].to;
        if(lca[x][0]==y) continue;
        lca[y][0]=x;                                       //小技巧:lca[x][0]即为x的父亲~~(向上跳2^0=1不就是父节点嘛)
        f[y][0]=e[i].data;
        d[y]=d[x]+1;
        dfs(y);                                            //再以这个节点为根建子树【这里真的用得到嘛??】
    }
}
int ask(int x,int y)                                             //询问,也是关键
{                                                                        
    if(d[x]<d[y]) {int t=x;x=y;y=t;}                  //把x搞成深的点
    int k=d[x]-d[y],ans=0;
    for(int i=0;i<=NI;i++)
        if(k&(1<<i))                                      //若能跳就把x跳一跳
            ans+=f[x][i],                              //更新信息
            x=lca[x][i];
    for(int i=NI;i>=0;i--)                                  //不知道能不能正着循环,好像倒着优,反正记得倒着就好了
        if(lca[x][i]!=lca[y][i])                            //如果x跳2^i和y跳2^j没跳到一起就让他们跳
            ans+=f[x][i]+f[y][i],
            x=lca[x][i],y=lca[y][i];
    return ans+f[x][0]+f[y][0];                           //跳到LCA上去(每步跳的时候都要更新信息,而且要在跳之前更新信息哦~)
}
int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    build(1,2,a);
    build(1,3,b);                                                       //分别建1 2、1 3之间的边
    dfs(1);                                                                //以1为根建树
    printf("%d",ask(2,3));                                         //求解2 3到它们的LCA的距离和并输出
}

手软了

此处乃原版

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