Boring Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 698 Accepted Submission(s):
346
Problem Description
Number theory is interesting, while this problem is
boring.
Here is the problem. Given an integer sequence a1, a2, …, an, let S(i) = {j|1<=j<i, and aj is a multiple of ai}. If S(i) is not empty, let f(i) be the maximum integer in S(i); otherwise, f(i) = i. Now we define b i as af(i). Similarly, let T(i) = {j|i<j<=n, and aj is a multiple of ai}. If T(i) is not empty, let g(i) be the minimum integer in T(i); otherwise, g(i) = i. Now we define ci as ag(i). The boring sum of this sequence is defined as b1 * c1 + b2 * c2 + … + bn * cn.
Given an integer sequence, your task is to calculate its boring sum.
Here is the problem. Given an integer sequence a1, a2, …, an, let S(i) = {j|1<=j<i, and aj is a multiple of ai}. If S(i) is not empty, let f(i) be the maximum integer in S(i); otherwise, f(i) = i. Now we define b i as af(i). Similarly, let T(i) = {j|i<j<=n, and aj is a multiple of ai}. If T(i) is not empty, let g(i) be the minimum integer in T(i); otherwise, g(i) = i. Now we define ci as ag(i). The boring sum of this sequence is defined as b1 * c1 + b2 * c2 + … + bn * cn.
Given an integer sequence, your task is to calculate its boring sum.
Input
The input contains multiple test cases.
Each case consists of two lines. The first line contains an integer n (1<=n<=100000). The second line contains n integers a1, a2, …, an (1<= ai<=100000).
The input is terminated by n = 0.
Each case consists of two lines. The first line contains an integer n (1<=n<=100000). The second line contains n integers a1, a2, …, an (1<= ai<=100000).
The input is terminated by n = 0.
Output
Output the answer in a line.
Sample Input
5
1 4 2 3 9
0
Sample Output
136
Hint
In the sample, b1=1, c1=4, b2=4, c2=4, b3=4, c3=2, b4=3, c4=9, b5=9, c5=9, so b1 * c1 + b2 * c2 + … + b5 * c5 = 136.Author
SYSU
Source
1 /** 2 给出一个数列：a[i]，然后 3 b[i]：表示在 i 前面的项，如果有a[i]的倍数（要最靠近i的），那么b[i]就等于这个数，如果没有那么b[i] = a[i]; 4 c[i]：表示在 i 后面的项，如果有a[i]的倍数（要最靠近i的），那么c[i] 就等于这个数，如果没有那么c[i] = a[i]; 5 **/ 6 #include<iostream> 7 #include<stdio.h> 8 #include<cstring> 9 #include<cstdlib> 10 #include<vector> 11 using namespace std; 12 13 int a[100002],b[100002],c[100002]; 14 vector<int>Q[100002]; 15 int hash1[100002]; 16 int main() 17 { 18 int n; 19 int MAX,MIN,k; 20 for(int i=2;i<=100000;i++) 21 for(int j=i;j<=100000;j=j+i) 22 Q[i].push_back(j); 23 while(scanf("%d",&n)>0) 24 { 25 if(n==0)break; 26 for(int i=1; i<=n; i++){ 27 scanf("%d",&a[i]); 28 b[i] = c[i] = a[i]; 29 } 30 memset(hash1,0,sizeof(hash1)); 31 hash1[a[1]] = 1; 32 for(int i=2;i<=n;i++) 33 { 34 if(a[i]==1){ 35 b[i] = a[i-1]; 36 continue; 37 } 38 k = Q[a[i]].size(); 39 MAX = -1; 40 for(int j=0;j<k;j++) 41 if(hash1[Q[a[i]][j]]!=0 && MAX<hash1[Q[a[i]][j]]) 42 MAX = hash1[Q[a[i]][j]]; 43 44 if(MAX == -1); 45 else b[i] = a[MAX]; 46 hash1[a[i]] = i; 47 } 48 memset(hash1,0,sizeof(hash1)); 49 hash1[a[n]] = n; 50 for(int i=n-1;i>=1;i--) 51 { 52 if(a[i]==1) { c[i] = a[i+1]; continue;} 53 MIN = 111111111; 54 k = Q[a[i]].size(); 55 for(int j=0;j<k;j++) 56 if(hash1[Q[a[i]][j]]!=0 && MIN>hash1[Q[a[i]][j]]) 57 MIN = hash1[Q[a[i]][j]]; 58 if(MIN ==111111111 ); 59 else c[i] = a[MIN]; 60 hash1[a[i]] = i; 61 } 62 __int64 sum = 0; 63 for(int i=1;i<=n;i++) 64 sum = sum+((__int64)b[i])*c[i]; 65 printf("%I64d ",sum); 66 } 67 return 0; 68 }