ZOJ 3627 Treasure Hunt II

ac了。。好水的题。。
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;

#ifdef _WIN32
#define i64 __int64
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
#define i64 long long
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FFF(i,a)        for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b)      for( int i = (a) ; i >= (b) ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define pb              push_back
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define read            freopen("in.txt","r",stdin)
#define write           freopen("out.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-9;
const double pi = acos(-1.0);
const int maxn = 100011;

int n,p;
i64 a[maxn];
int m,t;
i64 s[maxn];
int l,r;
i64 ans;
i64 now,to;

int main()
{
    while(cin>>n>>p)
    {
        MM(a,0);
        MM(s,0);
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        s[0]=0;
        for(int i=1;i<=n;i++)
        {
            s[i]=s[i-1]+a[i];
        }
        cin>>m>>t;
        ans=0;
        if(m/2>=t)
        {
            l = max(p-t,1);
            r = min(p+t,n);
            ans=s[r]-s[l-1];
            cout<<ans<<endl;
            continue;
        }
        else if(m%2==0)
        {
            l = p-t;
            r = p+m/2;
            while(l<=r && l<=p-m/2 && r>0)
            {
                now = max(l,1);
                to = min(r,n);
                ans = max(ans,s[to]-s[now-1]);
                l+=2;
                r++;
            }
            l = p-m/2;
            r = p + t;
            int now,to;
            while(l<=r && l<=n && r>=p+m/2)
            {
                now = max(l,1);
                to = min(r,n);
                ans = max(ans,s[to]-s[now-1]);
                l-=1;
                r-=2;
            }
            cout<<ans<<endl;
            continue;
        }
        else
        {
            l = p - t;
            r = p + m/2;
            now = max(l,1);
            to = min(r,n);

补充：软件开发 , C++ ,