题解

1232 条题解

  • 13
    @ 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(

  • 6
    @ 2017-08-05 11:40:48
    #include <iostream>
    using namespace std;
    int main()
    {
        int a,b;
        cin>>a>>b;
        cout<<a+b;
        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
    @ 2020-10-17 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;
    }

  • 0
    @ 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;
    }
    
  • 0
    @ 2020-10-18 17:08:13

    最短代码挑战

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

  • 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;
    }
    

信息

ID
1000
难度
9
分类
(无)
标签
(无)
递交数
67036
已通过
25980
通过率
39%
被复制
79