标签:highlight main mat pre bsp line system with cow
Farmer John‘s N (1 <= N <= 100,000) cows, conveniently numbered 1..N, are once again standing in a row. Cow i has height H_i (1 <= H_i <= 1,000,000).
Each cow is looking to her left toward those with higher index numbers. We say that cow i ‘looks up‘ to cow j if i < j and H_i < H_j. For each cow i, FJ would like to know the index of the first cow in line looked up to by cow i.
Note: about 50% of the test data will have N <= 1,000.
约翰的N(1≤N≤10^5)头奶牛站成一排,奶牛i的身高是Hi(l≤Hi≤1,000,000).现在,每只奶牛都在向右看齐.对于奶牛i,如果奶牛j满足i<j且Hi<Hj,我们可以说奶牛i可以仰望奶牛j. 求出每只奶牛离她最近的仰望对象.
Input
输入格式:
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains the single integer: H_i
第 1 行输入 N,之后每行输入一个身高 H_i。
输出格式:
* Lines 1..N: Line i contains a single integer representing the smallest index of a cow up to which cow i looks. If no such cow exists, print 0.
共 N 行,按顺序每行输出一只奶牛的最近仰望对象,如果没有仰望对象,输出 0。
FJ has six cows of heights 3, 2, 6, 1, 1, and 2.
Cows 1 and 2 both look up to cow 3; cows 4 and 5 both look up to cow 6; and cows 3 and 6 do not look up to any cow.
【输入说明】6 头奶牛的身高分别为 3, 2, 6, 1, 1, 2.
【输出说明】奶牛#1,#2 仰望奶牛#3,奶牛#4,#5 仰望奶牛#6,奶牛#3 和#6 没有仰望对象。
【数据规模】
对于 20%的数据: 1≤N≤10;
对于 50%的数据: 1≤N≤1,000;
对于 100%的数据:1≤N≤100,000;1≤H_i≤1,000,000;
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
int s[100001],a[100001],n;
int main(){
scanf("%d",&n);
for(int IAKNOIP=1;IAKNOIP<=n;IAKNOIP++){
scanf("%d",&a[IAKNOIP]);
}
for(int j,i=n-1;i>=1;i--){
j=i+1;
while((a[i]>=a[j])&&(a[j]>0)){
j=s[j];
}
s[i]=j;
}
for(int IAKNOIP2018=1;IAKNOIP2018<=n;IAKNOIP2018++){
printf("%d\n",s[IAKNOIP2018]);
}
///*system("pause");*/
return 0;
}
标签:highlight main mat pre bsp line system with cow
原文地址:https://www.cnblogs.com/xiongchongwen/p/11192366.html