Codeforces Round #420 (Div. 2) B. Okabe and Banana Trees

题目大意

给出一次函数y=-x/m+b,它与x轴y轴围成平面中，每个点（x，y）贡献（x+y），在这个平面中画一个矩形，所围成点的最大价值和是多少。

题解

分别计算纵坐标减1后在范围外的点和新增加的点。（要用long long,比赛时被max函数坑wa了2发）。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int read(){    char ch=getchar();int f=0;    while(ch<'0'||ch>'9') ch=getchar();    while(ch>='0'&&ch<='9')    {f=(f<<1)+(f<<3)+ch-'0';    ch=getchar();}    return f;}long long maxx(long long x,long long y){    return x>y?x:y;}long long a[10005];long long maxn=0,last[10005];int main(){    int m,b;    cin>>m>>b;    for(int i=b;i>=0;i--)    {        a[i]=(b-i)*m;        //a[i]++;        //cout<<a[i]<<" ";    }//  cout<<endl;    maxn=((1+b)*b)/2;    last[b]=maxn;    for(int i=b-1;i>=0;i--)    {        last[i]=last[i+1]-((a[i+1]+1)*(i+1))-((1+a[i+1])*a[i+1])/2+(((a[i+1]+1+a[i])*(a[i]-a[i+1]))/2)*(i+1)+(((1+i)*i)/2)*(a[i]-a[i+1]);        //cout<<last[i+1]<<" "<<((a[i+1]+1)*(i+1))<<" "<<((1+a[i+1])*a[i+1])/2<<" "<<(((a[i+1]+1+a[i])*(a[i]-a[i+1]))/2)*i<<" "<<(((1+i)*i)/2)*(a[i]-a[i+1])<<endl;        maxn=maxx(last[i],maxn);    }    cout<<maxn;}

ps：还是大爷的代码好看。。。

#include<bits/stdc++.h>using namespace std;typedef long long LL;inline int rd(void) {    int x = 0, c = 0, f = 1;    for(;c<'0'||c>'9';c=getchar())f=c!='-';    for(;c>='0'&&c<='9';c=getchar())x=x*10+c-'0';    return f ? x : -x;}int m,b;int now,last;LL ans;LL cal(int n,int m){    return 1LL*(n+1)*(m+1)*(n+m)/2;}int main(){    m=rd(),b=rd();    for(int i=0;i<=b;i++) {        ans=max(ans,cal(i,(b-i)*m));    }    cout<<ans<<endl;}