【贪心】Codeforces Round #402 (Div. 2) C. Dishonest Sellers
2017-02-26 19:38
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按照b[i]-a[i],对物品从大到小排序,如果这个值大于零,肯定要立刻购买,倘若小于0了,但是没买够K个的话,也得立刻购买。
#include<cstdio> #include<algorithm> using namespace std; struct Point { int x,y; }a[200010]; int n,K,ans; bool cmp(const Point &a,const Point &b) { return a.y-a.x>b.y-b.x; } int main() { // freopen("c.in","r",stdin); scanf("%d%d",&n,&K); for(int i=1;i<=n;++i) scanf("%d",&a[i].x); for(int i=1;i<=n;++i) scanf("%d",&a[i].y); sort(a+1,a+n+1,cmp); int i; for(i=1;i<=n;++i) { if(i>K && a[i].y-a[i].x<0) break; ans+=a[i].x; } for(;i<=n;++i) ans+=a[i].y; printf("%d\n",ans); return 0; }
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