7月11日 bc总结

A题：暴力替换p，然后贪心求最大字段和。贪心求最大字段和的方法居然忘了。。。

now每次加上a[i]，如果now>ans，更新ans；如果now<0，now=0.也就是L移到下一个区间。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<string>
#include<math.h>
#include<cctype>
#define ll long long
#define REP(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define REPP(i,a,b,t) for(int (i)=(a);(i)<=(b);(i)+=(t))
#define rep(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)
#define repp(i,a,b,t) for(int (i)=(a);(i)>=(b);(i)-=(t))
#define PII pair<int,int>
#define fst first
#define snd second
#define MP make_pair
#define PB push_back
#define RI(x) scanf("%d",&(x))
#define RII(x,y) scanf("%d%d",&(x),&(y))
#define RIII(x,y,z) scanf("%d%d%d",&(x),&(y),&(z))
#define DRI(x) int (x);scanf("%d",&(x))
#define DRII(x,y) int (x),(y);scanf("%d%d",&(x),&(y))
#define DRIII(x,y,z) int (x),(y),(z);scanf("%d%d",&(x),&(y),&(z))
#define RS(x) scanf("%s",x)
#define RSS(x,y) scanf("%s%s",x,y)
#define DRS(x) char x[maxn];scanf("%s",x)
#define DRSS(x,y) char x[maxn],y[maxn];scanf("%s%s",x,y)
#define MS0(a) memset((a),0,sizeof((a)))
#define MS1(a) memset((a),-1,sizeof((a)))
#define MS(a,b) memset((a),(b),sizeof((a)))
#define ALL(v) v.begin(),v.end()
#define SZ(v) (int)(v).size()

using namespace std;

const int maxn=1000100;
const int INF=(1<<29);
const double EPS=0.0000000001;
const double Pi=acos(-1.0);

ll a[maxn],p;
int n;

int main()
{
    DRI(T);
    while(T--){
        scanf("%d%I64d",&n,&p);
        REP(i,1,n) scanf("%I64d",&a[i]);
        ll ans=-INF;
        REP(i,1,n){
            ll t=a[i];
            a[i]=p;
            ll now=0;
            REP(j,1,n){
                now+=a[j];
                if(now>ans) ans=now;
                if(now<0) now=0;
            }
            a[i]=t;
        }
        cout<<ans<<endl;
    }
    return 0;
}

B题：给n个a[i]，m个b[j]；当a[i]>=b[j]时，可以用a[i]抵消b[i]，然后得分增加a[i]-b[i]。每个a[i]和b[j]只能用一次，求最大得分。

由于答案一定是k个a[i]抵消k个b[j]的，而且显然，一定是用最大的k个a[i]抵消k个最小的b[j].

因此，a[i]从大到小遍历，b[j]从小到大遍历，这样可以保证a[i]最大，b[j]最小，当a[i]<b[j]时，直接break。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<string>
#include<math.h>
#include<cctype>
#define ll long long
#define REP(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define REPP(i,a,b,t) for(int (i)=(a);(i)<=(b);(i)+=(t))
#define rep(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)
#define repp(i,a,b,t) for(int (i)=(a);(i)>=(b);(i)-=(t))
#define PII pair<int,int>
#define fst first
#define snd second
#define MP make_pair
#define PB push_back
#define RI(x) scanf("%d",&(x))
#define RII(x,y) scanf("%d%d",&(x),&(y))
#define RIII(x,y,z) scanf("%d%d%d",&(x),&(y),&(z))
#define DRI(x) int (x);scanf("%d",&(x))
#define DRII(x,y) int (x),(y);scanf("%d%d",&(x),&(y))
#define DRIII(x,y,z) int (x),(y),(z);scanf("%d%d",&(x),&(y),&(z))
#define RS(x) scanf("%s",x)
#define RSS(x,y) scanf("%s%s",x,y)
#define DRS(x) char x[maxn];scanf("%s",x)
#define DRSS(x,y) char x[maxn],y[maxn];scanf("%s%s",x,y)
#define MS0(a) memset((a),0,sizeof((a)))
#define MS1(a) memset((a),-1,sizeof((a)))
#define MS(a,b) memset((a),(b),sizeof((a)))
#define ALL(v) v.begin(),v.end()
#define SZ(v) (int)(v).size()

using namespace std;

const int maxn=1000100;
const int INF=(1<<29);
const double EPS=0.0000000001;
const double Pi=acos(-1.0);

int n,m;
vector<ll> A,B;
ll a,b;

int cmp(ll a,ll b)
{
    return a>b;
}

int main()
{
    DRI(T);
    while(T--){
        RII(n,m);
        A.clear();B.clear();
        REP(i,1,n) scanf("%I64d",&a),A.PB(a);
        REP(i,1,m) scanf("%I64d",&b),B.PB(b);
        sort(ALL(A),cmp);
        sort(ALL(B));
        int ia=0,ib=0;
        ll ans=0;
        for(;ia<n&&ib<m;){
            ll tmp=A[ia]-B[ib];
            if(tmp>0) ans+=tmp,ia++,ib++;
            else break;
        }
        cout<<ans<<endl;
    }
    return 0;
}