Codeforces Round 374 (Div 2)D Maxim and Array 【贪心】
2016-10-18 12:44
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【题意】
n个数,我们可以做k次操作,每次选定一个数,+x或-x,在这种条件下,让你求出怎么操作,可以使得乘积最小
【分析】
显然,如果负数个数为奇数,我们只需要使得所有数符号不变的条件下,绝对值尽可能大(从绝对值较小者开始递增)
如果负数个数为偶数,我们要先使得一个绝对值最小的数变号(非负->负 或 负->非负),然后再做上面的操作
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 2e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n, K, X;
LL a
;
struct node
{
LL x;
int sg;
int p;
bool operator < (const node &b)const
{
if (abs(x) != abs(b.x))return abs(x) < abs(b.x);//使得绝对值小的排在前面
return p < b.p;
}
};
set< node >sot;
int main()
{
while(~scanf("%d%d%d",&n,&K,&X))
{
sot.clear();
int neg = 0;
for (int i = 1; i <= n; ++i)
{
int x; scanf("%d", &x);
int sg = (x >= 0 ? 1 : -1);
if (x < 0)++neg;
sot.insert({ x, sg, i });
}
if (neg % 2 == 0)//负数个数为偶数,这种情况下,我们要变号
{
while (K)
{
--K;
LL x = sot.begin()->x;
int sg = sot.begin()->sg;
int p = sot.begin()->p;
sot.erase(sot.begin());
if ((x - X * sg) * sg <= 0)
{
sot.insert({ x - X * sg, -sg, p });
break;
}
else
{
sot.insert({ x - X * sg, sg, p });
}
}
}
while (K)
{
--K;
LL x = sot.begin()->x;
int sg = sot.begin()->sg;
int p = sot.begin()->p;
sot.erase(sot.begin());
sot.insert({ x + X * sg, sg, p });
}
for (auto it : sot)a[it.p] = it.x;
for (int i = 1; i <= n; ++i)printf("%lld ", a[i]);
puts("");
}
return 0;
}
n个数,我们可以做k次操作,每次选定一个数,+x或-x,在这种条件下,让你求出怎么操作,可以使得乘积最小
【分析】
显然,如果负数个数为奇数,我们只需要使得所有数符号不变的条件下,绝对值尽可能大(从绝对值较小者开始递增)
如果负数个数为偶数,我们要先使得一个绝对值最小的数变号(非负->负 或 负->非负),然后再做上面的操作
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 2e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n, K, X;
LL a
;
struct node
{
LL x;
int sg;
int p;
bool operator < (const node &b)const
{
if (abs(x) != abs(b.x))return abs(x) < abs(b.x);//使得绝对值小的排在前面
return p < b.p;
}
};
set< node >sot;
int main()
{
while(~scanf("%d%d%d",&n,&K,&X))
{
sot.clear();
int neg = 0;
for (int i = 1; i <= n; ++i)
{
int x; scanf("%d", &x);
int sg = (x >= 0 ? 1 : -1);
if (x < 0)++neg;
sot.insert({ x, sg, i });
}
if (neg % 2 == 0)//负数个数为偶数,这种情况下,我们要变号
{
while (K)
{
--K;
LL x = sot.begin()->x;
int sg = sot.begin()->sg;
int p = sot.begin()->p;
sot.erase(sot.begin());
if ((x - X * sg) * sg <= 0)
{
sot.insert({ x - X * sg, -sg, p });
break;
}
else
{
sot.insert({ x - X * sg, sg, p });
}
}
}
while (K)
{
--K;
LL x = sot.begin()->x;
int sg = sot.begin()->sg;
int p = sot.begin()->p;
sot.erase(sot.begin());
sot.insert({ x + X * sg, sg, p });
}
for (auto it : sot)a[it.p] = it.x;
for (int i = 1; i <= n; ++i)printf("%lld ", a[i]);
puts("");
}
return 0;
}
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