zoukankan      html  css  js  c++  java
  • HDU-4407-Sum(容斥原理)

    Problem Description
    XXX is puzzled with the question below: 

    1, 2, 3, ..., n (1<=n<=400000) are placed in a line. There are m (1<=m<=1000) operations of two kinds.

    Operation 1: among the x-th number to the y-th number (inclusive), get the sum of the numbers which are co-prime with p( 1 <=p <= 400000).
    Operation 2: change the x-th number to c( 1 <=c <= 400000).

    For each operation, XXX will spend a lot of time to treat it. So he wants to ask you to help him.
     

    Input
    There are several test cases.
    The first line in the input is an integer indicating the number of test cases.
    For each case, the first line begins with two integers --- the above mentioned n and m.
    Each the following m lines contains an operation.
    Operation 1 is in this format: "1 x y p". 
    Operation 2 is in this format: "2 x c".
     

    Output
    For each operation 1, output a single integer in one line representing the result.
     

    Sample Input
    1 3 3 2 2 3 1 1 3 4 1 2 3 6
     

    Sample Output
    7 0
     

    Source
     

    思路:m<=1000 和 数列初始状态为1,2,3,..n 是该题的突破口。

    对于每一次询问。先不考虑数字被改动了。那我们能够直接用求和公式把x到y之间的数字的和求出来,然后再减去那些不和p互质的数(用容斥原理),再对改动过的数进行特判就可以。


    #include <stdio.h>
    
    int num,idx[1005],val[1005],prime[10],p[40000];
    
    inline bool check(int x)
    {
        int i;
    
        for(i=0;i<num;i++) if(x%prime[i]==0) return 0;
    
        return 1;
    }
    
    int main()
    {
        int T,n,m,i,j,k,type,cnt,a,b,c,last,lxdcnt,lxdnum,l,r;
        long long ans;
    
        //把40W以内的素数预处理出来-----------------
        cnt=0;
    
        for(i=2;i<400000;i++)
        {
            for(j=2;j*j<=i;j++) if(i%j==0) break;
    
            if(j*j>i) p[cnt++]=i;
        }
        //------------------------------------------
    
        scanf("%d",&T);
    
        while(T--)
        {
            scanf("%d%d",&n,&m);
    
            cnt=0;
    
            for(i=1;i<=m;i++)
            {
                scanf("%d",&type);
    
                if(type==1)
                {
                    scanf("%d%d%d",&a,&b,&c);
    
                    ans=(long long)(a+b)*(b-a+1)/2;
    
                    num=0;//质因数的个数
    
                    //获取c的质因数-----------------
                    last=0;
    
                    while(c>1)
                    {
                        if(c%p[last]==0)
                        {
                            prime[num++]=p[last];
                            c/=p[last];
                            while(c%p[last]==0) c/=p[last];
                        }
    
                        last++;
                    }
                    //-------------------------------
    
                    //容斥原理-------------------------
                    for(j=1;j<(1<<num);j++)
                    {
                        lxdcnt=0;
                        lxdnum=1;
    
                        for(k=0;k<num;k++) if(j&(1<<k))
                        {
                            lxdcnt++;
                            lxdnum*=prime[k];
                        }
    
    
                        l=a/lxdnum*lxdnum;
                        if(l<a) l+=lxdnum;
                        r=b/lxdnum*lxdnum;
                        if(r<l) continue;
    
                        if(lxdcnt&1) ans-=(long long)(l+r)*((r-l)/lxdnum+1)/2;
                        else ans+=(long long)(l+r)*((r-l)/lxdnum+1)/2;
                    }
                    //-----------------------------------
    
                    //对改动过的数字特殊推断-----------------------------------
                    for(j=0;j<cnt;j++)
                    {
                        if(idx[j]>=a && idx[j]<=b)
                        {
                            if(!check(idx[j]))
                            {
                                if(check(val[j])) ans+=val[j];
                            }
                            else
                            {
                                if(check(val[j])) ans=ans+val[j]-idx[j];
                                else ans-=idx[j];
                            }
                        }
                    }
                    //--------------------------------------------------------
    
                    printf("%I64d
    ",ans);
                }
                else
                {
                    scanf("%d%d",&a,&b);
    
                    for(j=0;j<cnt;j++) if(idx[j]==a)//注意,改动的点可能之前已被改动过
                    {
                        val[j]=b;
                        break;
                    }
    
                    if(j==cnt)
                    {
                        idx[cnt]=a;
                        val[cnt++]=b;
                    }
                }
            }
        }
    }
  • 相关阅读:
    T450的Fn lock
    移民,不应该是走投无路后的选择
    门槛低的行业看天赋,门槛高的行业看毅力
    个人是时代的一朵浪花
    转载:XPath基本语法
    爪哇国新游记之三十四----Dom4j的XPath操作
    常去的论坛今天两个传统行业的坛友要下岗了
    异常中要了解的Throwable类中的几个方法
    感觉JVM的默认异常处理不够好,既然不好那我们就自己来处理异常呗!那么如何自己处理异常呢?
    JVM对异常的默认处理方案
  • 原文地址:https://www.cnblogs.com/yxwkf/p/5211264.html
Copyright © 2011-2022 走看看