bestcoder#32 1002 哈希+后缀数组
bestcoder#32 1002 哈希+后缀数组
Negative and Positive (NP)
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1161 Accepted Submission(s): 59
Problem Description
When given an array (a0,a1,a2,⋯an−1) and an integer K, you are expected to judge whether there is a pair (i,j)(0≤i≤j<n) which makes that NP−sum(i,j) equals to K true. Here NP−sum(i,j)=ai−ai+1+ai+2+⋯+(−1)j−iaj
Input
Multi test cases. In the first line of the input file there is an integer T indicates the number of test cases.
In the next 2∗T lines, it will list the data for each test case.
Each case occupies two lines, the first line contain two integers n and K which are mentioned above.
The second line contain (a0,a1,a2,⋯an−1)separated by exact one space.
[Technical Specification]
All input items are integers.
0<T≤25,1≤n≤1000000,−1000000000≤ai≤1000000000,−1000000000≤K≤1000000000
In the next 2∗T lines, it will list the data for each test case.
Each case occupies two lines, the first line contain two integers n and K which are mentioned above.
The second line contain (a0,a1,a2,⋯an−1)separated by exact one space.
[Technical Specification]
All input items are integers.
0<T≤25,1≤n≤1000000,−1000000000≤ai≤1000000000,−1000000000≤K≤1000000000
Output
For each case,the output should occupies exactly one line. The output format is Case #id: ans, here id is the data number starting from 1; ans is “Yes.” or “No.” (without quote) according to whether you can find (i,j) which makes PN−sum(i,j) equals to K.
See the sample for more details.
See the sample for more details.
Sample Input
2
1 1
1
2 1
-1 0
Sample Output
Case #1: Yes.
Case #2: No.
题意:查找是否存在sum(i,j)==K
思路:构造前缀和d[],对sum(i,j)=d[j]-d[i-1],从n到1遍历i-1,插入d[i-1],从哈希表中查找是否存在d[i-1]+K;
注意这里不从1到n遍历j是因为项的正负是由i-1决定,所以比较方便;当然从n开始遍历j更容易理解,但不好处理正负号
题目还卡了STL的set,更坑的是还卡了scanf,我也因此学会了手写hash和读入外挂,很好的一道题
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<ctype.h> #include<algorithm> #include<cmath> #include<set> using namespace std; const int maxn=1000100; const int HM=100100; int n,K; int a[maxn]; long long d[maxn]; struct Hash { long long date; Hash *next; };Hash h[HM]; void insert(long long date,Hash *h) { int t=abs(date)%HM; Hash *pre=&h[t],*p=pre->next; while(p!=NULL){ if(p->date=date) return; pre=p; p=p->next; } p=(Hash*)malloc(sizeof(Hash)); p->date=date;p->next=NULL; pre->next=p; } bool find(long long date,Hash *h) { int t=abs(date)%HM; Hash *pre=&h[t],*p=pre->next; while(p!=NULL){ if(p->date==date) return true; pre=p; p=p->next; } return false; } inline int read() { char c=getchar(); while(c!='-'&&!isdigit(c)) c=getchar(); int f=0,tag=1; if(c=='-'){ tag=-1; f=getchar()-'0'; } else f=c-'0'; while(isdigit(c=getchar())) f=f*10+c-'0'; return f*tag; } int main() { int T=read(); int tag=1; while(T--){ n=read();K=read(); d[0]=0; for(int i=1;i<=n;i++){ a[i]=read(); if(i&1) d[i]=d[i-1]+a[i]; else d[i]=d[i-1]-a[i]; } for(int i=0;i<HM;i++) h[i].next=NULL; bool flag=0; for(int i=n;i>=0;i--){ insert(d[i],h); if(i&1){ if(find(d[i]-K,h)) flag=1; } else if(find(d[i]+K,h)) flag=1; if(flag) break; } printf("Case #%d: ",tag++); if(flag) printf("Yes.\n"); else printf("No.\n"); } return 0; }
没有AC不了的题,只有不努力的ACMER!