【CodeForces 803B】Distances to Zero(模擬)
阿新 • • 發佈:2018-12-24
B. Distances to Zero
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
題目大意:n個元素,求每個數到離它最近的0的距離
思路:模擬瞎暴力,記錄相鄰兩個0的位置,依次做比較
You are given the array of integer numbersa0, a1, ..., an - 1. For each element find the distance to the nearest zero (to the element which equals to zero). There is at least one zero element in the given array.
Input
The first line contains integern(1 ≤ n ≤ 2·105)
— length of the arraya. The second line contains integer elements of the array separated by
single spaces ( - 109 ≤ ai ≤ 109).
Print the sequenced0, d1, ..., dn - 1, wherediis the difference of indices betweeniand nearestjsuch thataj = 0. It is possible thati = j.
Examples input9 2 1 0 3 0 0 3 2 4output
2 1 0 1 0 0 1 2 3input
5 0 1 2 3 4output
0 1 2 3 4input
7 5 6 0 1 -2 3 4output
2 1 0 1 2 3 4
#include <bits/stdc++.h> #define manx 200005 #define INF 0x3f3f3f3f using namespace std; int main() { int a[manx],ans[manx],n; while(~scanf("%d",&n)){ memset(a,0,sizeof(a)); memset(ans,0,sizeof(ans)); int k[2],t=0; //k記錄相鄰兩個0的位置 k[0]=k[1]=INF; for (int i=0; i<n; i++) scanf("%d",&a[i]); for (int i=0; i<n+1; i++){ if (!a[i]){ ans[i]=0; if (i!=n) k[t%2]=i; int f; if (k[(t+1)%2]==INF) f=0; else f=k[(t+1)%2]; for (int j=f; j<i; j++) ans[j]=min(abs(k[0]-j),abs(k[1]-j)); t++; } } for (int i=0; i<n; i++) printf("%d ",ans[i]); printf("\n"); } return 0; }