25 条题解
-
3
#include<bits/stdc++.h> using namespace std; int main() { string a1,b1; int a[10000]={},b[10000]={},c[10000]={}; cin >> a1 >> b1; int lena=a1.size(); int lenb=b1.size(); for(int i=1;i<=lena;i++) { a[i]=a1[lena-i]-'0'; } for(int i=1;i<=lenb;i++) { b[i]=b1[lenb-i]-'0'; } int lenc=1; while(lenc<=lena||lenc<=lenb) { c[lenc]+=a[lenc]+b[lenc]; if(c[lenc]>9) { c[lenc]-=10; c[lenc+1]++; } lenc++; } if(c[lenc]==0) { lenc--; } for(int i=lenc;i>=1;i--) { cout << c[i]; } return 0; } -
3
dingyizhen LV 7 @ 2025-10-25 9:34:12
#include<iostream> using namespace std; int main(){ int a,b; cin>>a>>b; cout<<a+b; return 0; } -
2
权威·忘本 @ 2025-5-11 9:37:50
权威
#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
yangshihuan
LV 7
@ 2024-7-26 10:58:29
A+B Problem题解
新用户强烈建议阅读此帖
首先我们要理清思路
1.需要用到什么样的头文件？
2.用什么样的数据范围？
3.思路是什么？
首先题目中的数据范围是1≤a,b≤10^6, 而int 的范围是-2147483648-2147483647 正合题意，所以数据类型可以用int
话不多说，直接上代码
#include<iostream>//导入头文件，iostream里面是标准输入输出流（我说的什么？） using namespace std;//使用标准命名空间 int main(){//主函数，程序的入口 int a,b;//创建a,b两个整型变量 cin>>a>>b;//输入 a , b 两个变量 cout<<a+b; //输出a+b的内容 return 0; }本蒟蒻发的第一篇题解，请多多支持喵~~
-
-1
sng2025102 LV 5 @ 2025-7-6 23:16:54
#include
using namespace std;
int main()
{
int a; int b; cin>>a>>b; cout<<a+b; return 0;}
-
-1
```cpp #include <bits/stdc++.h> using namespace std; #define LL long long const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; string a1 , b1; int a[500] , b[500] , c[500]; int main() { cin >> a1 >> b1; int lena = a1.size(); int lenb = b1.size(); for ( int i = 0 ; i < lena ; i++ ) { a[ lena - i ] = a1[i] - '0'; } for ( int i = 0 ; i < lenb ; i++ ) { b[ lenb - i ] = b1[i] - '0'; } int lenc = 1 , x = 0; while ( lenc <= lena || lenc <= lenb ) { c[lenc] = a[lenc] + b[lenc] + x; x = c[lenc] / 10; c[lenc] = c[lenc] % 10; lenc++; } if ( x > 0 ) { c[lenc] = x; } else { lenc--; } for ( int i = lenc ; i >= 1 ; i-- ) { cout << c[i]; } cout << endl; return 0; } //菜鸟驿站 //老六专属 -
-1
2024sng008 LV 8 @ 2025-1-23 11:13:08
#include<iostream> using namespace std; int main() { int a,b; cin>>a>>b; cout<<a+b; } -
-2
mengqingyu
LV 10
@ 2024-11-16 16:21:16
#include<iostream> using namespace std; int main(){ int a,b,c; cin>>a>>b; c=a+b; cout<<c; } -
-3
#A+B problem {
{ #include//头文件，使用的工具库为iostream using namespace std;//命名空间 int main()//主函数 //先问大家几个问题： //1.我们如何获取a与b？ //2.我们如何进行对a于b的运算? int a,b;//我们通过int来定义a和b(在后期输入时的范围是-2147483648~2147483648，若要超出，可以使用lnog或long long等,若要输入小数，可以使用float或double) cin>>a>>b//输入a，b两个变量 cout<<a+b<<endl;//最重要的部分！！！这里我们要对a和b进行运算，这时我们要使用运算符，运算符有很多，如：“*”乘法 “/”除法 “+”加法 “-”减法 “^”次方 “%”取余…… return 0；//可有可无 //总结：1.a与b通过int定义，且不同的类型变量有不同规定取值范围； //2.我们使用运算符进行两个变量的运算
-
-3
```cpp #include <bits/stdc++.h> using namespace std; #define LL long long const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; string a1 , b1; int a[500] , b[500] , c[500]; int main() { cin >> a1 >> b1; int lena = a1.size(); int lenb = b1.size(); for ( int i = 0 ; i < lena ; i++ ) { a[ lena - i ] = a1[i] - '0'; } for ( int i = 0 ; i < lenb ; i++ ) { b[ lenb - i ] = b1[i] - '0'; } int lenc = 1 , x = 0; while ( lenc <= lena || lenc <= lenb ) { c[lenc] = a[lenc] + b[lenc] + x; x = c[lenc] / 10; c[lenc] = c[lenc] % 10; lenc++; } if ( x > 0 ) { c[lenc] = x; } else { lenc--; } for ( int i = lenc ; i >= 1 ; i-- ) { cout << c[i]; } cout << endl; return 0; } //菜鸟驿站 //老六专属 -
-4
#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; }
