touxi @ 2021-11-05 19:29:41
#include<bits/stdc++.h>
using namespace std;
long long a,b;
long long sb(long long x)
{
if (x==1) return 1;
else return sb(x/2)+sb(x-x/2);
}
int main()
{
cin>>a>>b;
cout<<sb(a)+sb(b);
return 0;
}怎么优化
by matianchen @ 2021-11-07 13:22:23
怎么优化\
直接
#include<bits/stdc++.h>;
using namespace std;
int main(){
int a,b;
cin>>a>>b;
cout<<a+b;
return 0;
}就行了
by IGUNarding33 @ 2021-11-07 22:26:44
咱不能简单点吗...
by rz54cyq54088 @ 2021-11-17 13:41:06
这么简单的题被你搞的花里胡哨的
by felixesintot @ 2021-11-27 11:58:21
雾） 本题的恶搞题解到此为止。
为防止新人受到误导，不再接受新的此类题解。
以前的保留不会删除，但请不要再提交。
啊这。。。。。
by Terrysong @ 2021-12-07 22:44:38
这不是递归吗？
by codeLJH114514 @ 2021-12-15 19:49:49
只有一个数二分个辣子啊！
在我印象中不应该是两个数么？
真要二分：
#include <iostream>
using namespace std;
int l, r;
int a, b;
int main() {
cin >> a >> b;
l = -2000000000, r = 2000000000;
while (l <= r) {
int m = (l + r) / 2;
if (a + b < m) {
r = m - 1;
} else if (a + b == m) {
cout << m;
exit(0);
} else {
l = m + 1;
}
}
return 0;
}或者（递归版）
#include <iostream>
using namespace std;
int l, r;
int a, b;
int search(int l, int r) {
int m = (l + r) / 2;
if (a + b < m) {
return search(l, m - 1);
} else if (a + b > m) {
return search(m + 1, r);
} else {
return m;
}
}
int main() {
cin >> a >> b;
l = -2000000000, r = 2000000000;
cout << search(l, r);
return 0;
}by Dream1234 @ 2021-12-22 20:56:11
你想啥呢？为什么你不直接输出a+b???
代码 上！
#include<iostream>
using namespace std;
int main()
{
int a,b;
cin>>a>>b;//输入两个数
cout<<a+b;//输出他们的和
return 0;//愉快的结束
}by Roy_Yu @ 2021-12-24 18:51:29
#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;
}by xuzhihao1 @ 2022-01-25 13:48:45
二分查找？？？
by strcmp @ 2022-02-16 15:49:43
来自2022的问候
跟我以前蒟蒻的时候写快速幂一样（现在是大蒟蒻了