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);    }}