1153

1153 - 无影的神之右手

Time Limit：4s Memory Limit：512MByte

Submissions：261Solved：31

DESCRIPTION

觉不觉得这几个图很有毒啊？
其实这几个不算很有毒吧~
由乃懒得写题面了，反正写了也没人看，所以直接说题意吧~
给你一个序列a，每次查询一个区间[l,r]的乘积的约数个数mod 19260817

INPUT

第一行两个数n,m第二行n个数表示序列a后面m行每行两个数l,r表示查询区间[l,r]

OUTPUT

对于每个查询输出答案

SAMPLE INPUT

5 564 2 18 9 1001 52 42 31 43 4

SAMPLE OUTPUT

1651594510

HINT

n , m <= 100000 , a[i] <= 1000000

SOLUTION

“玲珑杯”ACM比赛 Round #20

解题思路：我们首先对每个数进行素数分解，大于1000的因子最多只有一个，小于1000的因子我们暴力记录，然后对于大于1000的因子用莫队处理就好。

#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>#include<cstring>#include<vector>using namespace std;using namespace std;typedef long long LL;inline bool scan_d(int &num){        char in;bool IsN=false;        in=getchar();        if(in==EOF) return false;        while(in!='-'&&(in<'0'||in>'9')) in=getchar();        if(in=='-'){ IsN=true;num=0;}        else num=in-'0';        while(in=getchar(),in>='0'&&in<='9'){                num*=10,num+=in-'0';        }        if(IsN) num=-num;        return true;}const int maxn = 100000 + 10;const LL mod = 19260817;int n, m;LL s0;int num[1000005];int a[maxn];int b[maxn];//大于1000的质因子LL term[maxn];int L, R;bool valid[10005];int prime[10005];int dp[maxn][200];//dp[i][j]前i个数,质因子为primep[j]的个数int inv[100005];int pos[maxn];int tot;int judge;int block;struct query{    int l, r;    int id;    int loc;    bool operator <(const query &res) const    {        if(loc == res.loc) return r < res.r;        else return loc < res.loc;    }}Query[maxn];LL quick_pow(LL x, LL i){    LL ans = 1;    x %= mod;    if(x == 1) return 1;    while(i)    {        if(i&1) ans = (ans * x) % mod;        i >>= 1;        x = (x * x) % mod;    }    return ans;}void getPrime(){    memset(valid, true, sizeof(valid));    tot = 0;    for(int i = 2; i <= 10000; i++)    {        if(valid[i])        {            prime[++tot] = i;        }        for(int j = 1; j <= tot && prime[j] * i <= 10000; j++)        {            valid[i * prime[j]] = false;            if(i % prime[j] == 0) break;        }    }    for(int i = 1; i <= tot; i++)    {        if(prime[i] > 1000)        {            judge = i - 1;            break;        }    }    for(int i = 1; i < 100004; i++)    {        inv[i] = (int)quick_pow((LL)i, mod - 2);    }}void init(){    L = 1;    R = 1;    memset(num, 0, sizeof(num));    memset(dp, 0, sizeof(dp));    memset(b, 0, sizeof(b));    block = sqrt(n);    s0 = 1;}void add(int x){    LL temp = (LL)inv[num[x] + 1];    s0 = (s0 * temp) % mod;    num[x]++;    s0 = (s0 * (num[x] + 1)) % mod;}void sub(int x){    LL temp = (LL)inv[num[x] + 1];    s0 = (s0 * temp) % mod;    num[x]--;    s0 = (s0 * (num[x] + 1)) % mod;}int main(){    //freopen("C:\\Users\\creator\\Desktop\\A2.in","r",stdin) ;    //freopen("C:\\Users\\creator\\Desktop\\out.txt","w",stdout) ;    getPrime();    //cout<<"judge == "<<judge<<endl;    while(~scanf("%d%d", &n, &m))    {        //printf("n == %d m == %d\n", n, m);        init();        for(int i = 1; i <= n; i++)        {            //scanf("%d", &a[i]);            scan_d(a[i]);            pos[i] = i / block;        }        for(int i = 1; i <= n; i++)        {            for(int j = 1; j <= judge; j++)            {                int temp = 0;                while(a[i] % prime[j] == 0)                {                    temp++;                    a[i] /= prime[j];                }                dp[i][j] = dp[i - 1][j] + temp;            }            if(a[i] != 1) b[i] = a[i];        }        for(int i = 1; i <= m; i++)        {            //scanf("%d%d", &Query[i].l, &Query[i].r);            scan_d(Query[i].l);            scan_d(Query[i].r);            Query[i].id = i;            Query[i].loc = pos[Query[i].l];        }        sort(Query + 1, Query + m + 1);        if(b[1] != 0) add(b[1]);        for(int i = 1; i <= m; i++)        {            int l = Query[i].l;            int r = Query[i].r;            while(R < r)            {                R++;                if(b[R])                {                    add(b[R]);                }            }            while(L > l)            {                L--;                if(b[L])                {                    add(b[L]);                }            }            while(L < l)            {                if(b[L])                {                    sub(b[L]);                }                L++;            }            while(R > r)            {                if(b[R])                {                    sub(b[R]);                }                R--;            }            LL re = 1;            for(int j = 1; j <= judge; j++)            {                LL d = (LL)(dp[r][j] - dp[l - 1][j]);                if(d) re = (re * (d + 1)) % mod;            }            re = (re * s0) % mod;            term[Query[i].id] = re;        }        for(int i = 1; i <= m; i++)        {            printf("%lld\n", term[i]);        }    }    return 0;}