UESTC 1269-ZhangYu Speech
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ZhangYu Speech
Time Limit: 3000/1000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others)
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
for each query:
- ZhangYu brother choose an index
x from1 ton . - For all indices
y (y <x ) calculate the differenceb y =a x −a y . - Then ZhangYu brother calculate
B 1 ,the sum of all by which are greater than0 , and scfkcf calculateB 2 , the sum of all by which are less than0 .
if
Input
The first line contains two integers
Output
For each of
Sample input and output
10 30324152397147
Next timeKeep some distance from meI agree
Hint
It's better to use "scanf" instead of "cin" in your code.
Source
题目大意:
给出有n个数字的字符串,m表示有多少个比赛回合,给出一个x则会就是第几个数,用这个数将他前面的数都减一遍,当差大于零的时候加到b1上,当小于零时加到b2上,最后都取绝对值,比较b1和b2的大小。
这个题主要注意超时,事先可以化简这个式子,得到公式,就是(x-1)*(map[x-1]-'0')-前x-1项和。所以只需要在运算前保留一下各个位置的前几项和就可以节约很多时间。
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char map[10000000];int wo6[1000000],wo6defeiqi[1000000];int main(){int n,m;while(scanf("%d%d",&n,&m)!=EOF){scanf("%s",map);int i,j;for(i=1;i<=n;i++){wo6[i]=map[i-1]-'0';}for(i=1;i<=n;i++){wo6defeiqi[i]=wo6[i]+wo6defeiqi[i-1];}while(m--){int axiba;int xx;scanf("%d",&xx);axiba=(xx)*(map[xx-1]-'0')-wo6defeiqi[xx];if(axiba>0)printf("Keep some distance from me\n");else if(axiba==0)printf("Next time\n");else if(axiba<0)printf("I agree\n");}}return 0;}
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