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Boring Sum（hdu4961）hash
Boring Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 814 Accepted Submission(s): 390
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.题意：
给出n个数的数列a，bi的取值为在1 <= j < i之间如果存在aj % ai == 0，
则取最大下标的值赋给bi，如果不存在，则bi = ai；ci的取值为在i < j <= n之间
如果存在aj % ai == 0，则取最小下标值赋给bi，如果不存在，则ci = ai。
求b1 * c1 + b2 * c2 + ... + bn * cn的和。
思路：
如果直接暴力的话一定会超时，所以我们可以开一个vis数组来记录每一个值
所对应的最大的下标是多少。即每查找ai，分解出ai的质因子，更新vis数组
#include<stdio.h>#include<string.h>#define ll __int64#define maxn 100000+5#define mem(x) memset(x,0,sizeof(x))ll a[maxn],b[maxn],c[maxn],sum,n;ll vis[maxn];//a[i]的下标iint main(){ ll i,j,k,temp; while(scanf("%I64d",&n),n) { mem(b); mem(c); mem(vis); for(i=1;i<=n;i++) scanf("%I64d",&a[i]); vis[a[1]]=1; for(i=2;i<=n;i++) { for(j=1;j*j<=a[i];j++) { if(a[i]%j!=0) continue;//取质因子// printf("i=%I64d j=%I64d a[i]=%I64d ",i,j,a[i]); if(vis[j]!=0)// { b[vis[j]]=a[i];// printf("b[vis[j]]=%I64d vis[j]=%I64d ",b[vis[j]],vis[j]); vis[j]=0; } temp=a[i]/j; // printf("temp=%I64d ",temp); if(vis[temp]!=0)//更新 { b[vis[temp]]=a[i]; // printf("b[vis[temp]]=%I64d vis[temp]=%I64d",b[vis[temp]],vis[temp]); vis[temp]=0; } // printf("\n"); } vis[a[i]]=i; } for(i=1;i<=n;i++) if(b[i]==0) b[i]=a[i]; mem(vis); vis[a[n]]=n; for(i=n-1;i>=1;i--) { for(j=1;j*j<=a[i];j++) { if(a[i]%j!=0) continue;//取质因子 if(vis[j]!=0)// { c[vis[j]]=a[i]; vis[j]=0; } temp=a[i]/j; if(vis[temp]!=0)//更新 { c[vis[temp]]=a[i]; vis[temp]=0; } } vis[a[i]]=i; } for(i=1;i<=n;i++) if(c[i]==0) c[i]=a[i]; // for(i=1;i<=n;i++) // printf("b:%I64d\tc:%I64d\n",b[i],c[i]); sum=0; for(i=1;i<=n;i++) sum+=b[i]*c[i]; printf("%I64d\n",sum); } return 0;}
Boring Sum（hdu4961）hash
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