UESTC 1269 ZhangYu Speech 预处理、前缀和

ZhangYu Speech

Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)

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as we all know, ZhangYu(Octopus vulgaris) brother has a very famous speech - "Keep some distance from me". ZhangYu brother is so rich that 

everyone want to contact he, and scfkcf is one of them. One day , ZhangYu brother agreed with scfkcf to contact him if scfkcf could beat him. 

There are  n  digits(lets give them indices from  1  to  n  and name them  a1,a2...aN    ) and some queries.

for each query:

1. ZhangYu brother choose an index  x  from  1  to  n .
2. For all indices  y  (  y  <  x ) calculate the difference  by=ax−ay .
3. Then ZhangYu brother calculate  B1  ，the sum of all by which are greater than  0  , and scfkcf calculate  B2  , the sum of all by which are 
4. less than  0 .

if  B1>|B2|  , ZhangYu brother won and did not agree with scfkcf to contact him; else if  B1  is equals to  |B2|  , ZhangYu brother would ignore 

the result; else if  B1 <  |B2|  , ZhangYu brother lost and agreed with scfkcf to contact him.

Input

The first line contains two integers  n ,  m   (1≤n,m≤100000)  denoting the number of digits and number of queries. The second line 

contains  n  digits (without spaces)  a1,a2,...,an . (0≤ai≤9)    Each of next  m  lines contains single integer  x   (1≤x≤n)  denoting the 

index for current query.

Output

For each of  m  queries print "Keep some distance from me" if ZhangYu won, else print "Next time" if ZhangYu brother ignored the result, else 

print "I agree" if ZhangYu brother lost in a line - answer of the query.

Sample input and output

Sample Input	Sample Output
10 3 0324152397 1 4 7	Next time Keep some distance from me I agree

Hint

It's better to use "scanf" instead of "cin" in your code.

Source

第七届ACM趣味程序设计竞赛第四场（正式赛）B

My Solution

打表搞出前n项和；//像这样输入一次数据，然后n(n<1e6)次询问的，一般要预处理一下，来减少复杂度；

主要是找出x前的所有数的和与line[x-1]*x比较，如果前面的和大一点，说明负数和大；如果小，说明正数和大；相等，则相等；

这样省去了每次做差的麻烦；

#include <iostream>
#include <cstdio>
//#define LOCAL
const int maxn=100008;
char ch[maxn];
int line[maxn],sum[maxn];
using namespace std;

int main()
{
    #ifdef LOCAL
    freopen("a.txt","r",stdin);
    #endif // LOCAL
    int n,m,x,t;
    scanf("%d%d",&n,&m);
    scanf("%s",ch);
    //直接先转化过来，这样扫一遍就好了。不然要好多边临时的转化
    for(int i=0;i<n;i++){
        line[i]=ch[i]-'0';
    }
    //直接预处理出前i项和  前缀和
    t=0;
    for(int i=0;i<n;i++){
    	t+=line[i];
    	sum[i]=t;
    }
    //每次询问直接用sum[x-1];
    while(m--){
        scanf("%d",&x);

        //x1=ch[x-1]-'0';//cout<<x1<<" ";
        if(sum[x-1]-line[x-1]*(x)<0) printf("Keep some distance from me");
        else if(sum[x-1]-line[x-1]*(x)>0) printf("I agree");
        else if(sum[x-1]-line[x-1]*(x)==0)printf("Next time");
        if(m) printf("\n");
    }

    return 0;
}

谢谢