HDU-4961 Boring Sum(アナログ)
Boring Sum
http://acm.hdu.edu.cn/showproblem.php?pid=4961 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Problem Description
Number theory is interesting, while this problem is boring.
Here is the problem. Given an integer sequence a
1, a
2, …, a
n, let S(i) = {j|1<=jj is a multiple of a
i}. If S(i) is not empty, let f(i) be the maximum integer in S(i); otherwise, f(i) = i. Now we define bi as a
f(i). Similarly, let T(i) = {j|ij is a multiple of a
i}. If T(i) is not empty, let g(i) be the minimum integer in T(i); otherwise, g(i) = i. Now we define c
i as a
g(i). The boring sum of this sequence is defined as b
1 * c
1 + b
2 * c
2 + … + b
n * c
n.
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 a
1, a
2, …, a
n (1<= a
i<=100000).
The input is terminated by n = 0.
Output
Output the answer in a line.
Sample Input
Sample Output
また無限に正解に近づいて書かれていないので、1つの[1,i)の中でa[i]点倍数点の下の句読点の最値を維持しようと思っていたが、2つの最値は維持しなければならず、できなかった.の
まず1回スキャンして、メンテナンス[1,i)の中でa[i]点の倍数点の下の句読点の最大値で、そして直ちにb[i]を更新します;更に逆さまにスキャンして、メンテナンス(i,n)の中でa[i]点の倍数点の下の句読点の最小値で、そして直ちにc[i]を更新します
最後に乗じて答えを出す
http://acm.hdu.edu.cn/showproblem.php?pid=4961 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Problem Description
Number theory is interesting, while this problem is boring.
Here is the problem. Given an integer sequence a
1, a
2, …, a
n, let S(i) = {j|1<=j
i}. If S(i) is not empty, let f(i) be the maximum integer in S(i); otherwise, f(i) = i. Now we define bi as a
f(i). Similarly, let T(i) = {j|i
i}. If T(i) is not empty, let g(i) be the minimum integer in T(i); otherwise, g(i) = i. Now we define c
i as a
g(i). The boring sum of this sequence is defined as b
1 * c
1 + b
2 * c
2 + … + b
n * c
n.
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 a
1, a
2, …, a
n (1<= a
i<=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.
また無限に正解に近づいて書かれていないので、1つの[1,i)の中でa[i]点倍数点の下の句読点の最値を維持しようと思っていたが、2つの最値は維持しなければならず、できなかった.の
まず1回スキャンして、メンテナンス[1,i)の中でa[i]点の倍数点の下の句読点の最大値で、そして直ちにb[i]を更新します;更に逆さまにスキャンして、メンテナンス(i,n)の中でa[i]点の倍数点の下の句読点の最小値で、そして直ちにc[i]を更新します
最後に乗じて答えを出す
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int MAXN=100005;
int n,f,g;
int a[MAXN],b[MAXN],c[MAXN],indx[MAXN];
long long ans;
vector<int> factor[MAXN];//factor[i] i
int main() {
for(int j=1;j<=MAXN;++j)
for(int i=j;i<=MAXN;i+=j)
factor[i].push_back(j);
while(scanf("%d",&n),n!=0) {
memset(indx,0,sizeof(indx));
for(int i=1;i<=n;++i) {//indx[x] [1,i) a[i]
scanf("%d",a+i);
f=indx[a[i]]==0?i:indx[a[i]];
b[i]=a[f];
for(int j=0;j<factor[a[i]].size();++j)
indx[factor[a[i]][j]]=max(indx[factor[a[i]][j]],i);
}
memset(indx,0x3f,sizeof(indx));
for(int i=n;i>0;--i) {//indx[x] (i,n] a[i]
g=indx[a[i]]==0x3f3f3f3f?i:indx[a[i]];
c[i]=a[g];
for(int j=0;j<factor[a[i]].size();++j)
indx[factor[a[i]][j]]=min(indx[factor[a[i]][j]],i);
}
ans=0;
for(int i=1;i<=n;++i)
ans+=(long long)b[i]*c[i];
printf("%I64d
",ans);
}
return 0;
}