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HDU 5785 Function 【倍增】 (2016 ACM/ICPC Asia Regional Dalian Online)

Function

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 976    Accepted Submission(s): 375


Problem Description
The shorter, the simpler. With this problem, you should be convinced of this truth.
  
  You are given an array A of N postive integers, and M queries in the form (l,r). A function F(l,r) (1lrN) is defined as:
F(l,r)={AlF(l,r1) modArl=r;l<r.
You job is to calculate F(l,r), for each query (l,r).
 

 

Input
There are multiple test cases.
  
  The first line of input contains a integer T, indicating number of test cases, and T test cases follow. 
  
  For each test case, the first line contains an integer N(1N100000).
  The second line contains N space-separated positive integers: A1,,AN (0Ai109).
  The third line contains an integer M denoting the number of queries. 
  The following M lines each contain two integers l,r (1lrN), representing a query.
 

 

Output
For each query(l,r), output F(l,r) on one line.
 

 

Sample Input
132 3 311 3
 

 

Sample Output
2
 

 

Source
2016 ACM/ICPC Asia Regional Dalian Online
 

 

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题目链接:

  http://acm.hdu.edu.cn/showproblem.php?pid=5875

题目大意:

  N个数(N<=100000),M个询问,每次询问L,R,求F(L,R)。

  F(L,R)=F(L,R-1)%A[R] , L<R

     =A[L] , L=R

题目思路:

  【倍增】

  这题数据水到暴力居然过了。。比赛的时候一想会T就没写。

  首先这题实际是求A[L]%A[L+1]%A[L+2]...%A[R]的值。

  那么很容易想到A[L]右边比他大的数都是没有必要取模的,变为找L右边第一个小于A[L]的取模

  模完后的结果缩小,再次寻找当前位置右边第一个小于此时的值的继续取模。可以证明这样的取模次数为log2N(100000取模后最大为49999)

  暴力的做法就是预处理的时候直接从后往前用单调栈求出位置X右边第一个小于A[X]的位置Y,next[X]=Y。查询的时候用next跳着模。

  当然这是不够的,因为可以构造出递减序列使得实际操作次数仍为N。

  用倍增的思想,对于位置i右侧的最长下降子序列,next[i][j]表示位置i右边第2j个的位置,预处理出每个位置的next数组

  每次做的时候尽量远跳,直到找到next[j][0]恰好小于当前的值z,则z对A[next[j][0]]取模,移动左标记,继续查找,直到超出R或者没有更小的数。

 

倍增:

技术分享
  1 //  2 //by coolxxx  3 //#include<bits/stdc++.h>  4 #include<iostream>  5 #include<algorithm>  6 #include<string>  7 #include<iomanip>  8 #include<map>  9 #include<stack> 10 #include<queue> 11 #include<set> 12 #include<bitset> 13 #include<memory.h> 14 #include<time.h> 15 #include<stdio.h> 16 #include<stdlib.h> 17 #include<string.h> 18 //#include<stdbool.h> 19 #include<math.h> 20 #define min(a,b) ((a)<(b)?(a):(b)) 21 #define max(a,b) ((a)>(b)?(a):(b)) 22 #define abs(a) ((a)>0?(a):(-(a))) 23 #define lowbit(a) (a&(-a)) 24 #define sqr(a) ((a)*(a)) 25 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b)) 26 #define mem(a,b) memset(a,b,sizeof(a)) 27 #define eps (1e-10) 28 #define J 10000 29 #define mod 1000000007 30 #define MAX 0x7f7f7f7f 31 #define PI 3.14159265358979323 32 #pragma comment(linker,"/STACK:1024000000,1024000000") 33 #define N 100004 34 #define M 18 35 using namespace std; 36 typedef long long LL; 37 double anss; 38 LL aans; 39 int cas,cass; 40 int n,m,lll,ans; 41 int a[N],s[N]; 42 int nextt[N][M]; 43 void init() 44 { 45     int i,j,k,top; 46     s[top=1]=n; 47     for(i=0;i<M;i++)nextt[n][i]=n+1; 48     for(i=n-1;i;i--) 49     { 50         while(top && a[i]<a[s[top]])top--; 51         if(!top)nextt[i][0]=n+1; 52         else nextt[i][0]=s[top]; 53         s[++top]=i; 54     } 55     for(j=1;j<M;j++) 56     { 57         for(i=1;i<n;i++) 58         { 59             k=nextt[i][j-1]; 60             nextt[i][j]=nextt[k][j-1]; 61         } 62     } 63 } 64 int work(int l,int r) 65 { 66     int i,j,z=a[l]; 67     if(l==r)return z; 68     while(l<r) 69     { 70         if(!z)return z; 71         for(i=0;i<M-1;i++) 72         { 73             if(a[nextt[l][0]]<=z || nextt[l][i]>r)break; 74             if(a[nextt[l][i+1]]<=z || nextt[l][i+1]>r) 75             { 76                 l=nextt[l][i]; 77                 i=-1; 78                 continue; 79             } 80         } 81         if(i==M || nextt[l][i]>r || !a[nextt[l][i]])return z; 82         z%=a[nextt[l][0]];l=nextt[l][0]; 83     } 84     return z; 85 } 86 int main() 87 { 88     #ifndef ONLINE_JUDGE 89 //    freopen("1.txt","r",stdin); 90 //    freopen("2.txt","w",stdout); 91     #endif 92     int i,j,k; 93     int x,y,z; 94 //    init(); 95     for(scanf("%d",&cass);cass;cass--) 96 //    for(scanf("%d",&cas),cass=1;cass<=cas;cass++) 97 //    while(~scanf("%s",s)) 98 //    while(~scanf("%d",&n)) 99     {100         scanf("%d",&n);101         for(i=1;i<=n;i++)scanf("%d",a+i);102         init();103         scanf("%d",&m);104         for(i=1;i<=m;i++)105         {106             scanf("%d%d",&x,&y);107             printf("%d\n",work(x,y));108         }109     }110     return 0;111 }112 /*113 //114 115 //116 */
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暴力:

技术分享
 1 // 2 //by coolxxx 3 //#include<bits/stdc++.h> 4 #include<iostream> 5 #include<algorithm> 6 #include<string> 7 #include<iomanip> 8 #include<map> 9 #include<stack>10 #include<queue>11 #include<set>12 #include<bitset>13 #include<memory.h>14 #include<time.h>15 #include<stdio.h>16 #include<stdlib.h>17 #include<string.h>18 //#include<stdbool.h>19 #include<math.h>20 #define min(a,b) ((a)<(b)?(a):(b))21 #define max(a,b) ((a)>(b)?(a):(b))22 #define abs(a) ((a)>0?(a):(-(a)))23 #define lowbit(a) (a&(-a))24 #define sqr(a) ((a)*(a))25 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))26 #define mem(a,b) memset(a,b,sizeof(a))27 #define eps (1e-10)28 #define J 1000029 #define mod 100000000730 #define MAX 0x7f7f7f7f31 #define PI 3.1415926535897932332 #pragma comment(linker,"/STACK:1024000000,1024000000")33 #define N 10000434 using namespace std;35 typedef long long LL;36 double anss;37 LL aans;38 int cas,cass;39 int n,m,lll,ans;40 int a[N],pre[N],s[N];41 void print()42 {43     int i;44     for(i=1;i<=n;i++)45         printf("%d ",pre[i]);46     puts("");47 }48 int main()49 {50     #ifndef ONLINE_JUDGE51 //    freopen("1.txt","r",stdin);52 //    freopen("2.txt","w",stdout);53     #endif54     int i,j,k;55     int x,y,z,top,l,r;56 //    init();57     for(scanf("%d",&cass);cass;cass--)58 //    for(scanf("%d",&cas),cass=1;cass<=cas;cass++)59 //    while(~scanf("%s",s))60 //    while(~scanf("%d",&n))61     {62         scanf("%d",&n);63         for(i=1;i<=n;i++)64         {65             scanf("%d",&a[i]);66         }67         top=1;68         s[top]=n;pre[n]=n+1;69         for(i=n-1;i;i--)70         {71             while(top && a[i]<=a[s[top]])72                 top--;73             if(!top)pre[i]=n+1;74             else pre[i]=s[top];75             s[++top]=i;76         }77         scanf("%d",&m);78         for(i=1;i<=m;i++)79         {80             scanf("%d%d",&l,&r);81             x=a[l];82             for(j=pre[l];j<=r;j=pre[j])83                 x%=a[j];84             printf("%d\n",x);85         }86         //print();87     }88     return 0;89 }90 /*91 //92 93 //94 */
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HDU 5785 Function 【倍增】 (2016 ACM/ICPC Asia Regional Dalian Online)