HDU 5288 OO’s Sequence

题意： 给你一个序列， 有一个函数 F（L,R） 其中 ai 均不能 被 aL … aR整除的  函数值是这个ai个数

思路 ： 反过来求 满足这样的条件的 ai 的区间，然后求和

#include<iostream>
#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef __int64 LL;
const int maxn = 100002;
const LL MOD = 1e9 + 7;
vector<int> Cnt[10005];
int Left[maxn], Right[maxn];
int Num[maxn],Vis[maxn];
int Scan()
{
    int res = 0, ch, flag = 0;

    if((ch = getchar()) == '-')                //判断正负
        flag = 1;

    else if(ch >= '0' && ch <= '9')            //得到完整的数
        res = ch - '0';
    while((ch = getchar()) >= '0' && ch <= '9' )
        res = res * 10 + ch - '0';

    return flag ? -res : res;
}
void Init() {
    for(int i = 1; i <= 10005; ++i)
    {
        for(int j = 1; j <= i; ++j)
        if(i % j == 0) Cnt[i].push_back(j);
    }
}
int main() {
    int n;
    Init();
    while(scanf("%d",&n) != EOF) {
        for(int i = 0; i < n; ++i) Num[i] = Scan();
        memset(Left,-1,sizeof(Left));
        memset(Right,-1,sizeof(Right));
        memset(Vis,-1,sizeof(Vis));
        //GET LEFT
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < Cnt[Num[i]].size(); ++j) {
                int tmp = Cnt[Num[i]][j];
                if(Vis[tmp] != -1 && Num[i] % tmp == 0) {
                    if(Left[i] == -1) Left[i] = Vis[tmp] + 1;
                    else Left[i] = max(Left[i],Vis[tmp]+1);
                }
            }
            Vis[Num[i]] = i;
        }
        //GET RIGHT
        memset(Vis,-1,sizeof(Vis));
        for(int i = n-1; i >= 0; --i) {
            for(int j = 0; j < Cnt[Num[i]].size(); ++j) {
                int tmp = Cnt[Num[i]][j];
                if(Vis[tmp] != -1 && Num[i] % tmp == 0) {
                    if(Right[i] == -1) Right[i] = Vis[tmp] - 1;
                    else Right[i] = min(Right[i],Vis[tmp]-1);
                }
            }
            Vis[Num[i]] = i;
        }
        for(int i = 0; i < n; ++i) {
            if(Left[i] == -1) Left[i] = 0;
            if(Right[i] == -1) Right[i] = n-1;
        }
        LL ans = 0;
        for(LL i = 0; i < n; ++i) {
            LL L = i - Left[i] + 1;
            LL R = Right[i] - i + 1;
            ans = (ans + L * R) % MOD;
        }
        printf("%I64d\n",ans);
    }
}