1328 条题解
-
0
雁南 LV 3 @ 2006-04-25 16:54:46
var
a,b:longint;
begin
readln(a,b);
writeln(a+b);
end. -
0
cj19940723 LV 3 @ 2006-03-23 17:06:49
var
a,b:longint;
begin
readln(a,b);
writeln('A+B=',a+b);
end.
这样只要两个变量，节省了空间，加快了运行速度！ -
-1
#include<bits/stdc++.h> using namespace std; struct sb{ int a,b; }; int main(){ sb n; cin>>n.a>>n.b; cout<<n.a+n.b<<'\n'; }-
longyitongxue @ 2025-07-09 21:46:23
结构体类型名不文明，差评 \(+\) 举报
-
-
-1
一道很好的 树状数组 题。
#include <iostream> #include <string> #include <cstdio> using namespace std; struct Array { int *arr; int lowbit(int x) { return x & -x; } int getsum(int x) { //计算[1,x]区间中数列和 int ret=0; while(x>0) { ret+=arr[x]; x=x-lowbit(x); } return ret; } void plus(int x, int v) { //把第x个元素增加v while(x <= (int)sizeof(arr)) { arr[x]=arr[x]+v; x=x+lowbit(x); } } Array() {} Array(int length, int arr[]) { this->arr=new int[length+1]; for(int i=1; i<=length; i++) this->arr[i]=0; for(int i=1; i<=length; i++) plus(i, arr[i-1]); } ~Array()=default; }; int main() { Array a; int arr[2]; scanf("%d%d", &arr[0], &arr[1]); a=Array(2, arr); printf("%d\n", a.getsum(2)); return 0; } -
-1
beijingzhan LV 4 @ 2024-12-09 16:51:14
我有两篇题解：
给新手小白的：#include<iostream> int main(){ int a,b; scanf("%d %d",&a,&b); printf("%d",a+b); return 0; }这是给大佬的：
#include<iostream> #include<string> #include<cstring> #include<cstdio> using namespace std; const int N = 1005; struct bign { int len,s[N]; bign() { memset(s,0,sizeof(s)); len=1; } bign(int num) { *this=num; } bign(char *num) { *this=num; } bign operator =(int num) { char c[N]; sprintf(c,"%d",num); *this=c; return *this; } bign operator =(const char *num) { len=strlen(num); for (int i=0;i<len;i++) s[i]=num[len-1-i]-'0'; return *this; } string str() { string res=""; for (int i=0;i<len;i++) res=(char)(s[i]+'0')+res; return res; } void clean() { while (len>1&&!s[len-1]) len--; } bign operator +(const bign &b) { bign c; c.len=0; for (int i=0,g=0;g||i<len||i<b.len;i++) { int x=g; if (i<len) x+=s[i]; if (i<b.len) x+=b.s[i]; c.s[c.len++]=x%10; g=x/10; } return c; } bign operator -(const bign &b) { bign c; c.len=0; int x; for (int i=0,g=0;i<len;i++) { x=s[i]-g; if (i<b.len) x-=b.s[i]; if (x>=0) g=0; else{ x+=10; g=1; }; c.s[c.len++]=x; } c.clean(); return c; } bign operator *(const bign &b) { bign c; c.len=len+b.len; for (int i=0;i<len;i++) for (int j=0;j<b.len;j++) c.s[i+j]+=s[i]*b.s[j]; for (int i=0;i<c.len-1;i++) { c.s[i+1]+=c.s[i]/10; c.s[i]%=10; } c.clean(); return c; } bool operator <(const bign &b) { if (len!=b.len) return len<b.len; for (int i=len-1;i>=0;i--) if (s[i]!=b.s[i]) return s[i]<b.s[i]; return false; } bign operator +=(const bign &b) { *this=*this+b; return *this; } bign operator -=(const bign &b) { *this=*this-b; return *this; } }; istream& operator >>(istream &in,bign &x) { string s; in>>s; x=s.c_str(); return in; } ostream& operator <<(ostream &out,bign &x) { out<<x.str(); return out; } int main(){ bign a,b,c; ios::sync_with_stdio(false); cin>>a>>b; c=a+b; cout<<c<<endl; return 0; }希望这篇题解有助于大家！
-
-1
A + B问题实在是太难了，不过用LCT还是能解出来的，代码如下
#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;
} -
-1
一个蒟蒻 (nzy2011) LV 4 @ 2023-05-26 17:02:17
让评测姬爆炸的代码
#include<bits/stdc++.h> using namespace std; long long a[32768][32768]; int main(){ int x,y; cin>>x>>y; for(int i=1;i<=32767;i++) for(int j=1;j<=32767;j++) a[i][j] = i+j; cout<<a[x][y]; return 0; }
雁南
LV 3
@ 2006-04-25 16:54:46
信息
- ID
- 1000
- 难度
- 9
- 分类
- (无)
- 标签
- (无)
- 递交数
- 74849
- 已通过
- 28666
- 通过率
- 38%
- 被复制
- 253