2014 ACM-ICPC Vietnam National Second Round

B. Sum

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Write a program to compute the following sum S given a positive integer n:

, where  is the largest integer not greater than x.

Input

The input file consists of several datasets. The first line of the input file contains the number of datasets which is a positive integer and is not greater than 30. The following lines describe the datasets.

Each dataset contains a positive integer n (n ≤ 1012) written on a separate line.

Output

For each dataset, write in one line the remainder of the computed sum S divided by 106.

Sample test(s)

input

2
1
5

output

1
10

思路：发现[x/i]的不同的值很少， 于是可以分段进行。

D. Treasure Box

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Your team was exploring an ancient city. Suddenly you found an old scroll with 2 integer numbers N and K, which encrypts the secret code to open a treasure box. Considering a transformation on an integer X described as follows:

X = X + X mod 100,

the secret code can be obtained by applying the above-described transformation K times successively to N.

Input

The input file consists of several datasets. The first line of the input file contains the number of datasets which is a positive integer and is not greater than 500.

Each dataset has two space-separated positive integers N and K (1 ≤ N ≤ 109, 1 ≤ K ≤ 109) written on a single line.

Output

For each dataset, write on a single line the secret number decrypted from N and K.

Sample test(s)

input

2
31102014 2
10101 10

output

31102056
10324

思路: X MOD 100 存在循环节。 

 

E. ACM

time limit per test

9 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

An IT company puts out an advertisement to hire staff for a newly established office. They received n applications and organized an interview to select the best ones. They want to recruit candidates with both high level of expertise and good teamwork skill. Each candidate is assigned an ACM (Ability Coefficient of Multi-collectives) score that represents how the candidate meets the company’s internal selection criteria. Initially, everyone’s ACM score is 1.

Candidates are arranged into a round table of n seats, indexed from 1 to n. The first person sits next to the second person and the nthperson. For each interview question, candidates with indices from L and R form a group and register their collective answer to the system. If L ≤ R, the group consists of candidates at indices L, L + 1, L + 2, ..., R. If L > R, the group consists of candidates at indicesL, L + 1, ..., N, 1, ..., R. Depending on the answer, the ACM score of each group member is either multipled by X or divided by Y (in the later case, it is guaranteed that all ACM scores of the group are divisible by Y).

During the interview, the company may also request the system to output the product of the ACM scores of a group. This product could be a large number, so the system has to only output the value at modulo P. In summary, the system has to handle the following three types of queries:

• 0 L R P – compute the product of the ACM scores of all candidates from L to R, modulo P
• 1 L R X – the ACM score of each candidate from L to R is multiplied by X
• 2 L R Y – the ACM score of each candidate from L to R is divided by Y

For every query, we have 1 ≤ L, R ≤ N, 1 ≤ P ≤ 109 + 7, 1 ≤ X, Y ≤ 150.

Your task is to implement the system and output the computed products for every query of type 0.

Input

The input file consists of several datasets. The first line of the input file contains the number of datasets which is a positive integer and is not greater than 20. The following lines describe the datasets.

Each dataset comes in the following format:

• The first line contains 2 integers n, m where n is the number of candidates and m is the number of queries to be processed(1 ≤ n, m ≤ 50000).
• In the next m lines, the ith line contains the ith query.

Output

For each dataset, write out the corresponding outputs for queries of type 0 where each query output is on a separate line.

思路： 建立35课线段树，没课更新一个质因数的次幂。

#include <bits/stdc++.h>
#define LL long long

using namespace std;

struct node
{
    int l,r;
    LL sum,push;
};

const int X=150;
const int N=50010;

LL prime[35];
int pr;
node tree[35][N*4];

void init()
{
    pr=0;
    for (int i=2;i<=X;i++)
    {
        bool flag=true;
        for (int j=2;j<i;j++) if (i%j==0) flag=false;
        if (flag)
            prime[pr++]=i;
    }
}

void build(int idx, int t, int l, int r)
{
    tree[idx][t].l=l; tree[idx][t].r=r;
    tree[idx][t].sum=0; tree[idx][t].push=0;
    if (l==r) return;
    int mid=(l+r)/2;
    build(idx,t*2,l,mid);
    build(idx,t*2+1,mid+1,r);
}

void pushdown(int idx,int t)
{
    if (tree[idx][t].l!=tree[idx][t].r)
    {
        tree[idx][t*2].push+=tree[idx][t].push;
        tree[idx][t*2+1].push+=tree[idx][t].push;
        tree[idx][t*2].sum+=tree[idx][t].push*(tree[idx][t*2].r-tree[idx][t*2].l+1);
        tree[idx][t*2+1].sum+=tree[idx][t].push*(tree[idx][t*2+1].r-tree[idx][t*2+1].l+1);

        tree[idx][t].push=0;
    }
}

LL query(int idx, int t, int l, int r)
{
    if (l<=tree[idx][t].l&&tree[idx][t].r<=r)
    {
        return tree[idx][t].sum;
    }
    int mid=(tree[idx][t].l+tree[idx][t].r)/2;
    pushdown(idx,t);
    if (r<=mid)
        return query(idx,t*2,l,r);
    else
    if (l>mid)
        return query(idx,t*2+1,l,r);
    else
        return query(idx,t*2,l,mid)+query(idx,t*2+1,mid+1,r);
}

void update(int idx, int t, int l, int r, int x)
{
    if (l<=tree[idx][t].l&&tree[idx][t].r<=r)
    {
        tree[idx][t].push+=x;
        tree[idx][t].sum+=x*(tree[idx][t].r-tree[idx][t].l+1);
        return;
    }
    int mid=(tree[idx][t].l+tree[idx][t].r)/2;
    pushdown(idx,t);
    if (l<=mid)
        update(idx,t*2,l,r,x);
    if (r>mid)
        update(idx,t*2+1,l,r,x);

    tree[idx][t].sum=tree[idx][t*2].sum+tree[idx][t*2+1].sum;
}

LL multi(LL x, LL k,LL p)
{
    LL ans=1;
    x=x%p;
    while(k>0)
    {
        if(k%2==1) ans=(ans*x)%p;
        k=k/2;
        x=(x*x)%p;
    }
    return ans;
}

int main()
{
    int T,n,m,c,l,r;
    LL p;
    scanf("%d",&T);
    init();

    while(T--)
    {
        scanf("%d%d",&n,&m);
        for (int i=0;i<pr;i++) build(i,1,1,n);

        while(m--)
        {
            scanf("%d%d%d%I64d",&c,&l,&r,&p);
            if (c==0)
            {
                LL sum=1;
                for (int i=0;i<pr;i++)
                {
                    LL x=0;
                    if (l<=r) x=query(i,1,l,r); else x=query(i,1,l,n)+query(i,1,1,r);
                    sum=(sum*multi(prime[i],(LL)x,p))%p;
                }
                cout<<sum<<endl;
            }
            if (c==1)
            {
                for (int i=0;i<pr;i++) if (p%prime[i]==0)
                {
                    int x=0;
                    while(p%prime[i]==0) {x++; p=p/prime[i];}
                    if (l<=r) update(i,1,l,r,x); else {update(i,1,l,n,x); update(i,1,1,r,x);}
                }
            }
            if (c==2)
            {
                for (int i=0;i<pr;i++) if (p%prime[i]==0)
                {
                    int x=0;
                    while(p%prime[i]==0) {x++; p=p/prime[i];}
                    if (l<=r) update(i,1,l,r,-x); else {update(i,1,l,n,-x); update(i,1,1,r,-x);}
                }
            }
        }
    }
    return 0;
}

/*
2
6 5
0 1 5 1000000007
1 2 4 15
0 1 6 8704173
2 2 3 3
0 1 6 1000000007
6 6
1 1 4 20
1 2 6 15
0 1 6 9704331
2 3 6 5
0 1 4 1000000007
0 1 5 1000000007
*/

　　

转载于:https://www.cnblogs.com/wzb-hust/p/4655627.html