题解

1328 条题解

  • -1
    @ 2020-12-18 16:14:57
    #include <iostream>
    
    using namespace std;
    
    int main()
    {
        int a, b;
        cin >> a >> b;
        cout << a + b << endl;
    }
    
    
  • -1
    @ 2020-12-13 21:28:02

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

  • -1
    @ 2020-11-04 21:05:01

    蒟蒻刚学oier,不会什么大佬们的神仙操作,只能照着书上说的打下来

    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
        int n,m;
        scanf("%d%d",&n,&m);
        printf("%d",n+m);
        return 0;
    }
    
  • -1
    @ 2020-10-31 12:04:51

    这题是Vijos的经典啊……

    这里给出一些比较基础的A+B方法

    SPFA:

    #include<cstdio>
    using namespace std;
    int n,m,a,b,op,head[200009],next[200009],dis[200009],len[200009],v[200009],l,r,team[200009],pd[100009],u,v1,e;
    int lt(int x,int y,int z)
    {
        op++,v[op]=y;
        next[op]=head[x],head[x]=op,len[op]=z;
    }
    int SPFA(int s,int f)//SPFA……
    {
        for(int i=1;i<=200009;i++){dis[i]=999999999;}
        l=0,r=1,team[1]=s,pd[s]=1,dis[s]=0;
        while(l!=r)
        {
            l=(l+1)%90000,u=team[l],pd[u]=0,e=head[u];
            while(e!=0)
            {
                v1=v[e];
                if(dis[v1]>dis[u]+len[e])
                {
                    dis[v1]=dis[u]+len[e];
                    if(!pd[v1])
                    {
                        r=(r+1)%90000,
                        team[r]=v1,
                        pd[v1]=1;
                    }
                }
                e=next[e];
            } 
        }
        return dis[f];
    }
    int main()
    {
        scanf("%d%d",&a,&b);
        lt(1,2,a);lt(2,3,b);//1到2为a,2到3为b,1到3即为a+b……
        printf("%d",SPFA(1,3));
        return 0;
    }
    

    Floyd:

    #include<iostream>
    #include<cstring>
    using namespace std;
    long long n=3,a,b,dis[4][4];
    int main()
    {
        cin>>a>>b;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                dis[i][j]=2147483647;
            }
        }
        dis[1][2]=a,dis[2][3]=b;
        for(int k=1;k<=n;k++)
        {
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=n;j++)
                {
                    dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);//Floyd……
                }
            }
        }
        cout<<dis[1][3];
    }
    

    递归:

    #include<iostream>
    using namespace std;
    long long a,b,c;
    long long dg(long long a)
    {
        if(a<=5){return a;}//防超时……
        return (dg(a/2)+dg(a-a/2));
    }
    int main()
    {
        cin>>a>>b;
        c=dg(a)+dg(b);
        cout<<c;
    }
    

    高精:

    #include<iostream>
    #include<cstring>
    using namespace std;
    int main()
    {
        char a1[1000],b1[1000];
          int a[1000]={0},b[1000]={0},c[1000]={0},la,lb,lc,i,x;
          cin>>a1>>b1;
          la=strlen(a1);
          lb=strlen(b1);
          for(i=0;i<=la-1;i++){a[la-i]=a1[i]-48;}
        for(i=0;i<=lb-1;i++){b[lb-i]=b1[i]-48;}
          lc=1,x=0;
        while(lc<=la||lc<=lb){c[lc]=a[lc]+b[lc]+x,x=c[lc]/10,c[lc]%=10,lc++;}
        c[lc]=x;
        if(c[lc]==0){lc--;}
        for(i=lc;i>=1;i--){cout<<c[i];}
        cout<<endl;
        return 0;
    }
    

    压位高精:

    #include <cstdio>  
    #include <cstring>  
    #include <cstdlib>  
    #include <iostream>  
    #define p 8
    #define carry 100000000
    using namespace std;  
    const int Maxn=50001;  
    char s1[Maxn],s2[Maxn];  
    int a[Maxn],b[Maxn],ans[Maxn];  
    int change(char s[],int n[])   
    {  
        char temp[Maxn];   
        int len=strlen(s+1),cur=0;  
        while(len/p)
        {  
            strncpy(temp,s+len-p+1,p);
            n[++cur]=atoi(temp); 
            len-=p;
        }  
        if(len)
        {
            memset(temp,0,sizeof(temp));  
            strncpy(temp,s+1,len);  
            n[++cur]=atoi(temp);   
        }  
        return cur;
    }  
    int add(int a[],int b[],int c[],int l1,int l2)  
    {  
        int x=0,l3=max(l1,l2);  
        for(int i=1;i<=l3;i++)
        {  
            c[i]=a[i]+b[i]+x;  
            x=c[i]/carry;
            c[i]%=carry;  
        }  
        while(x>0){c[++l3]=x%10;x/=10;}  
        return l3;
    }  
    void print(int a[],int len)  
    {   
        printf("%d",a[len]);
        for(int i=len-1;i>=1;i--)printf("%0*d",p,a[i]);
        printf("\n");  
    }  
    int main()  
    {
        scanf("%s%s",s1+1,s2+1);
        int la=change(s1,a);
        int lb=change(s2,b);
        int len=add(a,b,ans,la,lb);    
        print(ans,len);
    }  
    
    
  • -1
    @ 2020-10-24 15:17:45

    #include<bits/stdc++.h>//万能头
    using namespace std;
    long long a,b;//int 也可以,提防大数据
    int main(){//主函数
    cin>>a>>b;
    printf("%d",a+b);//**cout<<a+b也行**但相对来说printf更快
    return 0;
    }

  • -1
    @ 2020-10-18 17:08:13

    最短代码挑战

    print sum(map(int, raw_input().split()))

  • -1
    @ 2020-10-17 19:50:39

    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
    @ 2020-09-12 17:07:16
    #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
    @ 2020-09-06 17:39:58

    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;
    }
    
  • -1
    @ 2020-09-04 22:03:28

    #include <iostream>

    using namespace std;

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

  • -1
    @ 2020-08-30 11:50:05

    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
    @ 2020-08-29 13:55:24

    #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

  • -1
    @ 2020-08-21 11:09:25

    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

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

  • -1
    @ 2020-08-17 20:22:37

    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
    @ 2020-07-24 14:07:12
    #include <iostream>
    
    using namespace std;
    
    int main()
    {
        int a, b;
        cin >> a >> b;
        cout << a + b << endl;
    }
    
  • -1
    @ 2020-07-14 16:10:58

    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
    @ 2020-07-11 16:31:22
    #include <iostream>
    using namespace std;
    int main() {
        int a,b;
        cin >> a >> b;
        cout <<0;
        return 0;
    }
    
  • -1
    @ 2020-07-03 21:13:07
    #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
    @ 2020-06-26 13:06:51

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

信息

ID
1000
难度
9
分类
(无)
标签
(无)
递交数
74844
已通过
28664
通过率
38%
被复制
253