标签:操作 first syn ++ using node nbsp algorithm struct
题意:n个数,k次操作,每次对任意一个数+x 或 -x,使得所有的数乘积最小,求最后的数组
思路:一共有3种情况,1、初始有奇数个负号, 2、初始有偶数个负号,但是可以转化成奇数个负号, 3、初始有偶数个负号,并且不能转化成奇数个负号;分好类之后贪心,如果有奇数个负号,那么每次选择绝对值小的数操作,使其绝对值变大,如果是偶数个,那么先对绝对值最小的数操作,使其符号变化,如果是负数那么+x,否则-x,变为奇数个负号之后就和前面一样了
AC代码:
#include "iostream" #include "iomanip" #include "string.h" #include "stack" #include "queue" #include "string" #include "vector" #include "set" #include "map" #include "algorithm" #include "stdio.h" #include "math.h" #pragma comment(linker, "/STACK:102400000,102400000") #define bug(x) cout<<x<<" "<<"UUUUU"<<endl; #define mem(a,x) memset(a,x,sizeof(a)) #define step(x) fixed<< setprecision(x)<< #define mp(x,y) make_pair(x,y) #define pb(x) push_back(x) #define ll long long #define endl ("\n") #define ft first #define sd second #define lrt (rt<<1) #define rrt (rt<<1|1) using namespace std; const ll mod=1e9+7; const ll INF = 1e18+1LL; const int inf = 1e9+1e8; const double PI=acos(-1.0); const int N=2e5+100; struct Node{ ll v, fv, id; void myabs(){ fv=abs(v); } bool friend operator<(Node a, Node b){ return a.fv>b.fv; } }a; priority_queue<Node> Q; ll n,f,x,k,ans[N]; int main(){ ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); cin>>n>>k>>x; for(int i=1; i<=n; ++i){ cin>>a.v; a.id=i, a.myabs();//a.fv=abs(a.v); Q.push(a); if(a.v<0) f++; } if(f&1){ while(k--){ Node now=Q.top(); Q.pop(); if(now.v<0) now.v-=x; else now.v+=x; now.myabs(); Q.push(now); } } else{ Node now=Q.top(); Q.pop(); while(k--){ int f1=now.v>=0?1:-1; if(now.v<0) now.v+=x; else now.v-=x; now.myabs(); int f2=now.v>=0?1:-1; if(f1+f2==0){ Q.push(now); break; } } if(k==-1){ Q.push(now); k=0; } while(k--){ now=Q.top(); Q.pop(); if(now.v<0) now.v-=x; else now.v+=x; now.myabs(); Q.push(now); } } int t=Q.size(); while(!Q.empty()){ ans[Q.top().id]=Q.top().v, Q.pop(); } for(int i=1; i<=t; ++i){ cout<<ans[i]<<" "; } return 0; }
标签:操作 first syn ++ using node nbsp algorithm struct
原文地址:http://www.cnblogs.com/max88888888/p/7643520.html