hdoj 5288 OO’s Sequence

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5288

 //*************头文件区*************
 #include<iostream>
 #include<cstdio>
 #include<vector>
 #define N 100010
 #define P 1000000007
 using namespace std;
 int n,tmp,i,j;
 int r[N], l[N], q[N], a[N];
 long long a1,a2,sum;
 vector<int> vec[10010];
 int main()
 {
     freopen("1001.in","r",stdin);
     freopen("m1001.out","w",stdout);
     while(~scanf("%d",&n)){
         for( i = 101; i <= 10000; ++i)
             vec[i].clear();
 //          清空容器
         for( i = 1; i <= n; ++i){
             scanf("%d",&a[i]);
             l[i] = 0;
             r[i] = n+1;
             if( a[i] > 100)
                 vec[a[i]].push_back(i);
 //          数据初始化，超过100的数据将下标保存到vec[a[i]]内
         }
         for( j = 1; j <= 100; ++j){
             tmp = 0;
 //          暂时保存下标
             for( i = 1;i <= n; ++i){
                 if( a[i] % j ==0 )
                     l[i] = max(tmp,l[i]);
                 if( a[i] == j)
                     tmp = i;
             }
             tmp = n+1;
             for( i = n;i >= 1; --i){
                 if( a[i] % j ==0)
                     r[i] = min(tmp,r[i]);
                 if( a[i] == j)
                     tmp = i;
             }
 //          更新以100以内的数为约数的数的左右限制
         }

         for( i = 101; i <= 10000; ++i)
             q[i] = vec[i].size() - 1;
         for( i = n; i >= 1; --i){
             if(a[i] > 100){
                 for( j = a[i]; j <=  10000; j = j+a[i]){
                     while((q[j] >= 0)&&(vec[j][q[j]]>i)){
                         l[vec[j][q[j]]] = max(l[vec[j][q[j]]],i);
                         if((q[j] > 0)&&(vec[j][q[j]-1]>i))
                             q[j]--;
                         else
                             break;
                     }
                 }
             }
         }
         for( i = 101; i <= 10000; ++i)
             q[i] = 0;
 //          vec下标从0开始？
         for( i = 1; i <= n; ++i)
             if(a[i] > 100){
                 for( j = a[i]; j <=  10000; j = j+a[i]){
                     while((q[j]<vec[j].size())&&(vec[j][q[j]]<i)){
                         r[vec[j][q[j]]] = min(r[vec[j][q[j]]],i);
                         if((q[j]<vec[j].size()-1)&&(vec[j][q[j]+1]<i))
                             q[j]++;
                         else
                             break;
                     }
                 }
             }

         sum=0;
         for (i=1;i<=n;i++)
         {
             a1=r[i]-i;
             a2=i-l[i];
             sum=(sum+a1*a2)%P;
 //            printf("%lld\n",a1);
         }
        printf("%I64d\n",sum);
     }
 }