【HDOJ】5288 OO’s Sequence

二分寻找对于指定pos的最左因数点和最右因数点。

 /* 5288 */
 #include <iostream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 const int mod = 1e9+;
 const int maxn = 1e5+;
 const int maxm = ;
 vi vc[maxm];
 vi fac[maxm];
 int a[maxn];

 void init() {
     int i, j, k;

     for (i=; i<maxm; ++i) {
         for (j=; j*j<=i; ++j) {
             if (i%j == ) {
                 fac[i].pb(j);
                 k = i / j;
                 if (k != j)
                     fac[i].pb(k);
             }
         }
     }
 }

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     int n;
     int l, r, tmp;
     vi::iterator iter;
     __int64 ans;

     init();
     while (scanf("%d", &n) != EOF) {
         rep(i, , n+) {
             scanf("%d", &a[i]);
             vc[a[i]].pb(i);
         }

         int sz;
         ans = ;

         rep(i, , n+) {
             l = ;
             r = n+;
             sz = SZ(fac[a[i]]);
             rep(j, , sz) {
                 tmp = fac[a[i]][j];
                 iter = upper_bound(all(vc[tmp]), i);
                 if (iter != vc[tmp].end()) {
                     r = min(r, *iter);
                 }
                 if (iter!=vc[tmp].begin()) {
                     --iter;
                     if (*iter != i)
                         l = max(l, *iter);
                     else if (iter != vc[tmp].begin()) {
                         --iter;
                         l = max(l, *iter);
                     }
                 }
             }
             #ifndef ONLINE_JUDGE
                 tmp = (r-i) * (i-l);
                 // printf("%d: l = %d, r = %d, c = %d\n", i, l, r, tmp);
             #endif
             ans = (ans + 1LL * (r-i) * (i-l) % mod) % mod;
         }

         printf("%I64d\n", ans);

         rep(i, , maxm)
             vc[i].clear();
     }

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }