题解

1246 条题解

  • 12
    @ 2018-05-01 16:13:03

    告诉你们什么叫做暴力的题解。

    #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
    
    • @ 2018-05-01 22:51:50

      #include<bits/stdc++.h>
      #define qi int main()
      #define yin
      {
      #define huo int a,b;
      #define fu cin>>a>>b;
      #define bi cout<<a+b;
      #define qu return 0;
      #define zhi
      }
      using namespace std;
      qi
      yin
      huo
      fu
      bi
      qu
      zhi

    • @ 2019-07-11 18:13:13

      学习了

    • @ 2020-05-18 20:07:53

      zzmg,jbl(

  • 3
    #include <stdio.h>
    #include <windows.h>
    #include <time.h>
    int main () {
        int a, b;
        scanf("%d %d", &a, &b);
        srand(time(NULL));
        for (;;) {
            int c = rand() % (a + b) + 1;
            if (c == a + b) {
                printf("%d", c);
                return 0;       
            }
        }
    }
    

    这题去年IOI最难的,正解随机数,无他解,其他的都是牛马题解!!!!
    虽然这是正解,但还是有可能TLE!!!!

  • 3
    @ 2017-08-05 11:40:48
    #include <iostream>
    using namespace std;
    int main()
    {
        int a,b;
        cin>>a>>b;
        cout<<a+b;
        return 0;
    }
    
  • 2

    上次的题解有点问题,但是通过我们IOI中国队120948572341241294812个夜晚的思考,终于研究出了该世纪难题的真正的正解,加了一些优化,已经可以攻破很多个点的TLE了,但仍有待改进
    CODE:

    #include <stdio.h>
    #include <windows.h>
    #include <time.h>
    
    #define N 100000
    #define o3 __attribute__((optimize("-O3")))
    #define r register
    
    #pragma GCC optimize(3)
    #pragma GCC optimize(2)
    #pragma GCC optimize("-Ofast")
    
    o3 int main () {
        int a, b, before[N], l = 0;
        bool bz = true;
        scanf("%d %d", &a, &b);
        srand(time(NULL));
        for (;;) {
            int c = rand() % (a + b) + 1;
            for (r int i = 1; i <= l; ++ i) {
                if (before[i] == c) {
                    bz = false;
                    break;
                }
            }
            if (bz == false) {
                bz = true;
                continue;
            }
            if (c == a + b and bz == true) {
                printf("%d", c);
                before[++ l] = c;
                return 0;       
            }
        }
    }
    

    希望有能力的牛马可以帮忙优化一下!!!!!!

  • 0
    @ 2021-07-12 09:53:30

    *

    *#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);
            connect(A,B);
            cut(A,B);
            connect(A,B);
        printf("%d\n",query(A,B)); 
        return 0;
    }*
    

    *

  • 0
    @ 2021-05-04 16:28:02
    #include <iostream>
    using namespace std;
    int main () {
            int a,b;
          cin >> a >> b;
          cout << a+b << endl;
    }
    
  • 0
    @ 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);
    }  
    
    
  • 0
    @ 2020-09-04 22:03:28

    #include <iostream>

    using namespace std;

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

  • 0
    @ 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;
    }
    
  • 0
    @ 2020-01-03 20:31:47
    #include <bits/stdc++.h>
    
    int main(int a, int b, int k)
    {
        if (k) scanf("%d%d", &a, &b);
        b == 0 ? printf("%d\n", a) : main(a ^ b, (a & b) << 1, 0);
    }
    

    main递归 + 模拟位运算

  • 0
    @ 2019-11-21 12:29:00

    216
    4165

  • 0
    @ 2019-11-20 13:03:52

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

    }

  • 0
    @ 2019-11-17 10:09:24

    作死ing

    #include<bits/stdc++.h>
    using namespace std;
    char s[10001],ss[10001];
    int a[10001],b[10001],c[10001],j;
    bool x=false;
    int main() {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        scanf("%s%s",s,ss);
        a[0]=strlen(s);
        b[0]=strlen(ss);
        for(int i=1; i<=a[0]; i++) a[i]=s[a[0]-i]-'0';
        for(int i=1; i<=b[0]; i++) b[i]=ss[b[0]-i]-'0';
        for(j=1; j<=max(a[0],b[0])+1; j++) {
            c[j]=a[j]+b[j];
            if(c[j]>=10) {
                c[j]%=10;
                a[j+1]++;
            }
        }
        c[0]=j;
        if(c[j+1]>0) c[0]++;
        for(int i=c[0]; i>=1; i--) {
            if(x==false&&c[i]==0) continue;
            x=true;
            cout<<c[i];
        }
        printf("\n");
        return 0;
    }
    
  • 0
    @ 2019-11-17 10:07:35

    本蒟蒻的第一篇题解

    结构体+选排

    #include<bits/stdc++.h>
    using namespace std;
    struct eee
    {
        char xibu,ganbu;
        string name;
        int pjcj,pycj,lws;
        long long RMB=0;
    };
    eee a[101];
    int n;
    long long sum=0;
    int main()
    {
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i].name;
            cin>>a[i].pjcj>>a[i].pycj;
            cin>>a[i].ganbu>>a[i].xibu;
            cin>>a[i].lws;
            if(a[i].lws>=1&&a[i].pjcj>=81) a[i].RMB+=8000;
            if(a[i].pjcj>85&&a[i].pycj>80) a[i].RMB+=4000;
            if(a[i].pjcj>90) a[i].RMB+=2000;
            if(a[i].pjcj>85&&a[i].xibu=='Y') a[i].RMB+=1000;
            if(a[i].ganbu=='Y'&&a[i].pycj>80) a[i].RMB+=850;
            sum+=a[i].RMB;
        }
        for(int i=1;i<=n-1;i++)
            for(int j=i+1;j<=n;j++)
                if(a[i].RMB<a[j].RMB)
                {
                    swap(a[i].ganbu,a[j].ganbu);
                    swap(a[i].lws,a[j].lws);
                    swap(a[i].name,a[j].name);
                    swap(a[i].pjcj,a[j].pjcj);
                    swap(a[i].pycj,a[j].pycj);
                    swap(a[i].RMB,a[j].RMB);
                    swap(a[i].xibu,a[j].xibu);
                }
        cout<<a[1].name<<endl;
        cout<<a[1].RMB<<endl;
        cout<<sum<<endl;
        return 0;
    }
    
  • 0
    @ 2019-11-13 21:30:54

    dalao们求一发模拟退火的题解OuO

  • 0
    @ 2019-10-26 14:17:06

    本题链接:https://vijos.org/p/1000
    JieKe08的第1篇题解!

    for beginners,特设此题,^_^

    #include<bits/stdc++.h>//多美妙的万能头
    using namespace std;
    int main()
    {
        int a,b;//定义变量
        cin>>a>>b;//输入
        cout<<a+b;//输出,最后的换行可有可无
        return 0;//题目给的代码没有这个,但考试最好加上 
    }
    

    感谢您花费1分钟阅读鄙人的题解!

  • 0
    @ 2019-09-25 20:22:56

    快读你们懂得吧
    cpp
    #include<bits/stdc++.h>
    using namespace std;
    int a,b;
    inline int read(){
    int ret=0,f=1;char ch=getchar();
    while(!isdigit(ch)) f=(ch=='-'?-f:f),ch=getchar();
    while(isdigit(ch)) ret=ret*10+ch-'0',ch=getchar();
    return ret*f;
    }
    int main(){
    a=read(),b=read();
    printf("%d\n",a+b);
    return 0;
    }

  • 0
    @ 2019-09-11 20:20:15
    //最简做法
    #include<cstdio>
    int main()
    {
        int a,b;
        scanf("%d%d",&a,&b);
        printf("%d",a+b);
        return 0;
    }
    
  • 0
    @ 2019-08-31 15:03:02
    #include <bits/stdc++.h>
    using namespace std;
    int main(){
        int a,b;
        cin>>a>>b;
        cout<<a+b;
        return 0;
    }
    
  • 0
    @ 2019-08-27 18:04:02

    Python3
    cpp
    a,b = map(int,input().split())
    print(a + b)

信息

ID
1000
难度
9
分类
(无)
标签
(无)
递交数
68700
已通过
26634
通过率
39%
被复制
91