题解 - Vijos

-1

最短代码挑战
print sum(map(int, raw_input().split()))

-1

python的解法，使用map函数才可以。如果使用
x=int(input())
y=int(input())
print(x+y)
这种简单的思维不行，递交出错。
要使用如下代码中：

x,y=map(int,input().split())
print(x+y)

-1

Vanyun LV 5 @ 4 年前

#include<bits/stdc++.h>
using namespace std;

int a,b,c;

signed main(){
    cin>>a>>b;
    c=a+b;
    cout<<c<<endl;
    return 0;
}

-1

Y_Kirito LV 4 @ 4 年前

新手都能学会的简易卡常a+b

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#include<iostream>
using namespace std;
inline int read()
{
    register int x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')
            f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=(x<<1)+(x<<3)+(ch^48);
        ch=getchar();
    }
    return x*f;
}
int main()
{
    register int a,b;
    a=read(),b=read();
    printf("%d",a+b);
    return 0;
}

红烧清蒸某邓吖 @ 4 年前

同是天涯优化人，相逢何必又相识

-1

张铭杨 LV 4 @ 4 年前

#include <iostream>

using namespace std;

int main()
{
int a, b;
cin >> a >> b;
cout << a + b << endl;
}

-1

louis110 LV 4 @ 4 年前

最简单的题？按题目模拟即可

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
int a, b;
int main()
{
    scanf("%d %d", &a, &b);
    printf("%d", a + b);
    return 0;
}

-1

Wj (hlbchenweiping) LV 6 @ 4 年前

#include<bits/stdc++.h>
#define gou int main()
#define li {
#define guo int a,b;
#define jia cin>>a>>b;
#define sheng cout<<a+b;
#define si return 0;
#define yi }
using namespace std;
gou
li
guo
jia
sheng
si
yi

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弑魔之眼 LV 6 @ 4 年前

C

#include <stdio.h>

int main() {
    int a,b;
    scanf("%d%d",&a,&b);
    printf("%d", a+b);
    return 0;
}

c++

#include <iostream>
#include <cstdio>

using namespace std;

int main() {
    int a,b;
    cin >> a >> b;
    cout << a+b;
    return 0;
}

-1

猜猜我大号是谁awa (WMY-2020) LV 4 @ 4 年前

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

-1

lirunning LV 4 @ 4 年前

哇！终于找到了一道那么难的LCT题目！！！
赶紧来写一个Link-Cut Tree题解！

#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

李昊谦 LV 5 @ 4 年前

#include <iostream>

using namespace std;

int main()
{
    int a, b;
    cin >> a >> b;
    cout << a + b << endl;
}

-1

limingyang LV 8 @ 4 年前

都正常点！
俗话说得好，装*遭**（光速逃
下面是正经A+B
C++语言

#include <bits/stdc++.h>
using namespace std;
int main(){
    int a, b;
    cin >> a >> b;
    cout << a+b << endl;
    return 0;
}

-1

洛谷用户_2020王茂钺 LV 9 @ 4 年前

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

-1

孙鹤鸣 (sunheming) LV 10 @ 4 年前

#Python 管你高精不高精，a=int(input())b=int(input())print(a+b)
a=int(input())
b=int(input())
print(a+b)

更666：
print(int(input())+int(input())

-1

houpingze (houpingze) LV 5 @ 4 年前

#include<iostream>
using namespace std;
int n,cnt,m;
int main(){
cin>>n>>m;
cout<<n+m;
}

-1

笑戎悵惘 LV 4 @ 4 年前

最小生成树最好了
```cpp
#include <cstdio>
#include <algorithm>
#define INF 2140000000
using namespace std;
struct tree{int x,y,t;}a[10];
bool cmp(const tree&a,const tree&b){return a.t<b.t;}
int f[11],i,j,k,n,m,x,y,t,ans;
int root(int x){if (f[x]==x) return x;f[x]=root(f[x]);return f[x];}
int main(){
for (i=1;i<=10;i++) f[i]=i;
for (i=1;i<=2;i++){
scanf("%d",&a[i].t);
a[i].x=i+1;a[i].y=1;k++;
}
a[++k].x=1;a[k].y=3,a[k].t=INF;
sort(a+1,a+1+k,cmp);
for (i=1;i<=k;i++){
// printf("%d %d %d %d\n",k,a[i].x,a[i].y,a[i].t);
x=root(a[i].x);y=root(a[i].y);
if (x!=y) f[x]=y,ans+=a[i].t;
}
printf("%d\n",ans);
}

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gsq20080915 (gsq20080915) LV 4 @ 4 年前

1

-1

gsq20080915 (gsq20080915) LV 4 @ 4 年前

再也简单不过的题了

#include<cstdio>
int main()
{
    int a,b;
    scanf("%d %d",&a,&b);
  printf("%d",a+b);
}

-1

CatWigle281 LV 7 @ 4 年前

各位，请不要误导新人。。。

Python代码：

s = input().split()
print(int(s[0])+int(s[1]))

-1

xhc2007a LV 4 @ 4 年前

递推即可

#include<stdio.h>
#include<stdbool.h>
int main(){
    int a[1005],b[1005],sum[1005];
    int x,y,i;
    bool flag=0;
    scanf("%d%d",&x,&y);
    for(i=1;i<=1000;i++)a[i]=x;
    for(i=1;i<=1000;i++)b[i]=y;
    for(i=1;i<=1000;i++)sum[i]=a[i]+b[i];
    for(i=1;i<=999;i++)
        if(sum[i]!=sum[i+1]){
            printf("Impossable!");
            flag=1;
            break;
        }
    if(!flag)printf("%d",sum[1]);
    return 0;
}

题解

1325 条题解

A+B Problem

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题解

1325 条题解

A+B Problem

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