Codeforces Round #374 (Div. 2) D. Maxim and Array 贪心+ 最小堆
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D. Maxim and Arraytime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputRecently Maxim has found an array of n integers, needed by no one. He immediately come up with idea of changing it: he invented positive integer x and decided to add or subtract it from arbitrary array elements. Formally, by applying single operation Maxim chooses integer i (1 ≤ i ≤ n) and replaces the i-th element of array ai either with ai + x or with ai - x. Please note that the operation may be applied more than once to the same position.
Maxim is a curious minimalis, thus he wants to know what is the minimum value that the product of all array elements (i.e.
) can reach, if Maxim would apply no more than k operations to it. Please help him in that.
InputThe first line of the input contains three integers n, k and x (1 ≤ n, k ≤ 200 000, 1 ≤ x ≤ 109) — the number of elements in the array, the maximum number of operations and the number invented by Maxim, respectively.
The second line contains n integers a1, a2, ..., an (
) — the elements of the array found by Maxim.
OutputPrint n integers b1, b2, ..., bn in the only line — the array elements after applying no more than k operations to the array. In particular,
should stay true for every 1 ≤ i ≤ n, but the product of all array elements should be minimum possible.
If there are multiple answers, print any of them.
Examplesinput5 3 15 4 3 5 2output5 4 3 5 -1input5 3 15 4 3 5 5output5 4 0 5 5input5 3 15 4 4 5 5output5 1 4 5 5input3 2 75 4 2output5 11 -5贪心，一开始的思路错了，一开始想先找出负数的个数，如果为偶数个则找到绝对值最小的那个将它变成异号，否则就不变。然后再找到绝对值最小的那个，将剩余的k全部加或者减在上面，就是这个地方有了问题，就比如3 2 3-2 4 6如果照着我的想法，就是 -8 4 6但是 -5 7 6明显更好，所以我想到了应该k的变化应该一次次进行，每次找到绝对值最小的那个进行变化。#include<iostream>#include<cstring>#include<cstdio>#include<vector>#include<queue>#include<algorithm>#include<cmath>#include<stack>const int inf=0x3f3f3f3f;typedef long long LL;using namespace std;const int MAXN=2e5+10;LL a[MAXN];LL b[MAXN];struct Node{ int id; LL value; Node(int i,LL v) { id=i; value=v; }};struct cmp{ bool operator()(Node a,Node b){ return a.value>b.value; }};int main(){ int n,k,x; while(cin>>n>>k>>x) { int sum=0; for(int i=1;i<=n;i++) { scanf("%I64d",&a[i]); b[i]=a[i]; if(a[i]<0) { sum++; b[i]*=-1LL; } } int h=0; if(sum%2==0) { LL MIN=b[1]; int index=1; for(int i=2;i<=n;i++) { if(MIN>b[i]) { MIN=b[i]; index=i; } } int t=k; if(MIN/x+1<=k) { b[index]-=LL(MIN/x+1)*LL(x); k-=(MIN/x+1); } else { b[index]-=LL(k*x); k=0; } if(a[index]<0) { a[index]+=LL(t-k)*LL(x); } else { a[index]-=LL(t-k)*LL(x); } } if(k) { for(int i=1;i<=n;i++) { b[i]=a[i]; if(b[i]<0) { b[i]*=(-1LL); } } priority_queue<Node,vector<Node>,cmp> Q; for(int i=1;i<=n;i++) { Q.push(Node(i,b[i])); } while(k) { Node u=Q.top(); Q.pop(); b[u.id]+=LL(x); Q.push(Node(u.id,b[u.id])); k--; } for(int i=1;i<=n;i++) { if(a[i]<0) { a[i]=b[i]*-1LL; } else { a[i]=b[i]; } } } printf("%I64d",a[1]); for(int i=2;i<=n;i++) { printf(" %I64d",a[i]); } cout<<endl; }}
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