【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。

  1 /* 5288 */
  2 #include <iostream>
  3 #include <string>
  4 #include <map>
  5 #include <queue>
  6 #include <set>
  7 #include <stack>
  8 #include <vector>
  9 #include <deque>
 10 #include <algorithm>
 11 #include <cstdio>
 12 #include <cmath>
 13 #include <ctime>
 14 #include <cstring>
 15 #include <climits>
 16 #include <cctype>
 17 #include <cassert>
 18 #include <functional>
 19 #include <iterator>
 20 #include <iomanip>
 21 using namespace std;
 22 //#pragma comment(linker,"/STACK:102400000,1024000")
 23
 24 #define sti                set<int>
 25 #define stpii            set<pair<int, int> >
 26 #define mpii            map<int,int>
 27 #define vi                vector<int>
 28 #define pii                pair<int,int>
 29 #define vpii            vector<pair<int,int> >
 30 #define rep(i, a, n)     for (int i=a;i<n;++i)
 31 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 32 #define clr                clear
 33 #define pb                 push_back
 34 #define mp                 make_pair
 35 #define fir                first
 36 #define sec                second
 37 #define all(x)             (x).begin(),(x).end()
 38 #define SZ(x)             ((int)(x).size())
 39 #define lson            l, mid, rt<<1
 40 #define rson            mid+1, r, rt<<1|1
 41
 42 const int mod = 1e9+7;
 43 const int maxn = 1e5+5;
 44 const int maxm = 10005;
 45 vi vc[maxm];
 46 vi fac[maxm];
 47 int a[maxn];
 48
 49 void init() {
 50     int i, j, k;
 51
 52     for (i=1; i<maxm; ++i) {
 53         for (j=1; j*j<=i; ++j) {
 54             if (i%j == 0) {
 55                 fac[i].pb(j);
 56                 k = i / j;
 57                 if (k != j)
 58                     fac[i].pb(k);
 59             }
 60         }
 61     }
 62 }
 63
 64 int main() {
 65     ios::sync_with_stdio(false);
 66     #ifndef ONLINE_JUDGE
 67         freopen("data.in", "r", stdin);
 68         freopen("data.out", "w", stdout);
 69     #endif
 70
 71     int n;
 72     int l, r, tmp;
 73     vi::iterator iter;
 74     __int64 ans;
 75
 76     init();
 77     while (scanf("%d", &n) != EOF) {
 78         rep(i, 1, n+1) {
 79             scanf("%d", &a[i]);
 80             vc[a[i]].pb(i);
 81         }
 82
 83         int sz;
 84         ans = 0;
 85
 86         rep(i, 1, n+1) {
 87             l = 0;
 88             r = n+1;
 89             sz = SZ(fac[a[i]]);
 90             rep(j, 0, sz) {
 91                 tmp = fac[a[i]][j];
 92                 iter = upper_bound(all(vc[tmp]), i);
 93                 if (iter != vc[tmp].end()) {
 94                     r = min(r, *iter);
 95                 }
 96                 if (iter!=vc[tmp].begin()) {
 97                     --iter;
 98                     if (*iter != i)
 99                         l = max(l, *iter);
100                     else if (iter != vc[tmp].begin()) {
101                         --iter;
102                         l = max(l, *iter);
103                     }
104                 }
105             }
106             #ifndef ONLINE_JUDGE
107                 tmp = (r-i) * (i-l);
108                 // printf("%d: l = %d, r = %d, c = %d\n", i, l, r, tmp);
109             #endif
110             ans = (ans + 1LL * (r-i) * (i-l) % mod) % mod;
111         }
112
113         printf("%I64d\n", ans);
114
115         rep(i, 0, maxm)
116             vc[i].clear();
117     }
118
119     #ifndef ONLINE_JUDGE
120         printf("time = %d.\n", (int)clock());
121     #endif
122
123     return 0;
124 }