【题目链接】:http://codeforces.com/contest/779/problem/C
【题意】
有n个商品;
打折前买和打折后买的价格不一样;
且必须有至少k个商品在打折前买;
问你买走全部n个商品最少需要多少钱;
【题解】
/*
每件商品先选择打折前和打折后里面价格较低的;
然后把价格全部加起来(我好像神经病写了个排序);
然后看看你选了几个第一种商品;
少选了,就选一些在打折后买的商品加上差价在打折前买;
把差价升序排一下,选择大于等于0的加上差价就是答案了
if (a[i]<b[i])
{
c[i].x = a[i];
c[i].id = 1;
}
else
{
//a[i]>=b[i]
c[i].x = b[i];
c[i].id = 2;
}
sort(c+1,c+1+n);
int ans = 0,num[3];
rep1(i,1,n)
ans+=c[i].x,num[c[i].id]++;
if (num[1]<k)
{
rep1(i,1,n)
c[i].x = b[i].x-a[i].x;
sort(c);
rep1(i,1,n)
if (c[i].x>=0)
{
if (num[1]<k)
{
ans+=c[i].x;
num[1]++;
}
else
break;
}
}
*/
【完整代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%lld",&x)
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };
const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };
const double pi = acos(-1.0);
const int N = 2e5+1000;
struct abc
{
int x, id;
};
int a[N], b[N],n ,k, num[3];
abc c[N];
bool cmp1(abc a, abc b)
{
return a.x < b.x;
}
int main()
{
//freopen("F:\rush.txt", "r", stdin);
rei(n), rei(k);
rep1(i, 1, n)
rei(a[i]);
rep1(i, 1, n)
rei(b[i]);
rep1(i,1,n)
if (a[i]<b[i])
{
c[i].x = a[i];
c[i].id = 1;
}
else
{
//a[i]>=b[i]
c[i].x = b[i];
c[i].id = 2;
}
sort(c + 1, c + 1 + n,cmp1);
int ans = 0;
rep1(i, 1, n)
ans += c[i].x, num[c[i].id]++;
if (num[1]<k)
{
rep1(i, 1, n)
c[i].x = a[i] - b[i];
sort(c+1,c+1+n,cmp1);
rep1(i, 1, n)
if (c[i].x >= 0)
{
if (num[1]<k)
{
ans += c[i].x;
num[1]++;
}
else
break;
}
}
printf("%d
", ans);
return 0;
}