poj 3484 Showstopper-白红宇

poj 3484 Showstopper

阅读量：6821 次

发布时间：2019-06-26

本文共 3201 字，大约阅读时间需要 10 分钟。

Showstopper

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2236		Accepted: 662

Description

Data-mining huge data sets can be a painful and long lasting process if we are not aware of tiny patterns existing within those data sets.

One reputable company has recently discovered a tiny bug in their hardware video processing solution and they are trying to create software workaround. To achieve maximum performance they use their chips in pairs and all data objects in memory should have even number of references. Under certain circumstances this rule became violated and exactly one data object is referred by odd number of references. They are ready to launch product and this is the only showstopper they have. They need YOU to help them resolve this critical issue in most efficient way.

Can you help them?

Input

Input file consists from multiple data sets separated by one or more empty lines.

Each data set represents a sequence of 32-bit (positive) integers (references) which are stored in compressed way.

Each line of input set consists from three single space separated 32-bit (positive) integers X Y Z and they represent following sequence of references: X, X+Z, X+2*Z, X+3*Z, …, X+K*Z, …(while (X+K*Z)<=Y).

Your task is to data-mine input data and for each set determine weather data were corrupted, which reference is occurring odd number of times, and count that reference.

Output

For each input data set you should print to standard output new line of text with either “no corruption” (low case) or two integers separated by single space (first one is reference that occurs odd number of times and second one is count of that reference).

Sample Input

1 10 12 10 11 10 11 10 11 10 14 4 11 5 16 10 1

Sample Output

1 1no corruption4 3

Source

/** @Author: Lyucheng* @Date:   2017-07-29 20:58:28* @Last Modified by:   Lyucheng* @Last Modified time: 2017-08-02 09:59:49*//* 题意：每次给你X,Y,Z表示产品序列 X,X+Z,X+2*Z,X+3*Z,...,X+k*Z(X+K*Z
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
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             #include 
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               #include 
               
                #include 
                
                 #include 
                 
                  #include 
                  
                   #define MAXN 1024#define MAXM 1000005#define MAX 0xffffffff#define MIN 0#define ULL unsigned long longusing namespace std;char str[MAXN];ULL X[MAXM],Y[MAXM],Z[MAXM];ULL res;int n;inline ULL getsum(ULL m){ ULL sum=0; for(int i=0;i
                   
                    =Y[i]){ sum+=(Y[i]-X[i])/Z[i]+1; }else if(m>=X[i]){ sum+=(m-X[i])/Z[i]+1; } } return sum;}inline void init(){ n=0; res=0;}int main(){ // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); while(gets(str)){ init(); if(strlen(str)>=2){ //不是空的 sscanf(str,"%I64d %I64d %I64d",&X[n],&Y[n],&Z[n]); res^=( (Y[n]-X[n])/Z[n]+1); n++; } while(gets(str)!=NULL){ if(strlen(str)>=2){ //不是空的 sscanf(str,"%I64d %I64d %I64d",&X[n],&Y[n],&Z[n]); res^=( (Y[n]-X[n])/Z[n]+1); n++; }else{ break; } }//读入完成 if(n==0){ continue; } if(!(res&1)){ //如果调用次数本身就是偶数 cout << "no corruption" << endl; continue; } ULL l=MIN,r=MAX,m; while(r-l>0){ m=(l+r)/2; if(getsum(m)%2==MIN){ //前缀和是偶数的 l=m+1; }else{ //前缀和是奇数的 r=m; } } cout<
                    
                     <<" "<
                     
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转载于:https://www.cnblogs.com/wuwangchuxin0924/p/7272335.html

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