邏輯思考題，個人感覺非常複雜，參考網路上解法 AC
很難用言語解釋 看 code 反而比較清楚
//
// GGGGGGGGGGGGG CCCCCCCCCCCCC AAA
// GGG::::::::::::G CCC::::::::::::C A:::A
// GG:::::::::::::::G CC:::::::::::::::C A:::::A
// G:::::GGGGGGGG::::G C:::::CCCCCCCC::::C A:::::::A
// G:::::G GGGGGG C:::::C CCCCCC A:::::::::A
//G:::::G C:::::C A:::::A:::::A
//G:::::G C:::::C A:::::A A:::::A
//G:::::G GGGGGGGGGGC:::::C A:::::A A:::::A
//G:::::G G::::::::GC:::::C A:::::A A:::::A
//G:::::G GGGGG::::GC:::::C A:::::AAAAAAAAA:::::A
//G:::::G G::::GC:::::C A:::::::::::::::::::::A
// G:::::G G::::G C:::::C CCCCCC A:::::AAAAAAAAAAAAA:::::A
// G:::::GGGGGGGG::::G C:::::CCCCCCCC::::C A:::::A A:::::A
// GG:::::::::::::::G CC:::::::::::::::C A:::::A A:::::A
// GGG::::::GGG:::G CCC::::::::::::C A:::::A A:::::A
// GGGGGG GGGG CCCCCCCCCCCCCAAAAAAA AAAAAAA
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
#include <ctime>
using namespace std;
#ifdef DEBUG
#define VAR(a,b) decltype(b) a=(b)
#define debug(...) printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#define gettime() end\_time=clock();printf("now running time is %.7f\\n",(float)(end\_time - start\_time)/CLOCKS\_PER\_SEC);
#else
#define VAR(a,b) \_\_typeof(b) a=(b)
#define debug(...)
#define gettime()
#endif
typedef unsigned int uint;
typedef long long int Int;
typedef unsigned long long int UInt;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\\n")
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()
#define De(x)printf(#x"%d\\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\
printf("%d ",dp\[htx\]\[hty\]);}Pln();}
#define M 3005
#define PII pair<int,int\>
#define PB push\_back
#define oo INT\_MAX
#define Set\_oo 0x3f
#define FOR(a,b) for(VAR(a,(b).begin());a!=(b).end();++a)
#define eps 1e-6
#define X first
#define Y second
#define gca printf("im here\\n");
clock\_t start\_time=clock(), end\_time;
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
int tmp=min(x,y);
return min(tmp,z);
}
int max3(int x,int y,int z){
int tmp=max(x,y);
return max(tmp,z);
}
int n;
int nr,nl;
int h\[M\];
int al\[M\],ar\[M\];
int ans;
int lm\[M\],rm\[M\];
void solve(){
int pre=0;
al\[0\]=0;
lm\[0\]=0;
for(int i=1;i<=nl;i++){
if(h\[i\]>=h\[pre\]){
al\[i\]=al\[i-1\]+h\[pre\]\*((i-pre)<<1);
lm\[i\]=pre=i;
}else{
al\[i\]=al\[i-1\];
lm\[i\]=pre;
}
}
pre=n;
ar\[n\]=0;
rm\[n\]=n;
for(int i=n-1;i>=nr;i--){
if(h\[i\]>=h\[pre\]){
ar\[i\]=ar\[i+1\]+h\[pre\]\*((pre-(i))<<1);
rm\[i\]=pre=i;
}else{
ar\[i\]=ar\[i+1\];
rm\[i\]=pre;
}
}
// debug("%d %d\\n",h\[lm\[nl\]\],h\[rm\[nr\]\]);
if(h\[lm\[nl\]\]<h\[rm\[nr\]\]){
// debug("ok\\n");
int i=nr;
bool fg=false;
int index=-1;
for(;h\[lm\[nl\]\]>=h\[i\];i++){
if(!fg&&h\[lm\[nl\]\]==h\[i\]){
fg=1;
index=i;
}
}
if(fg&&al\[lm\[nl\]\]<=(i-index)\*(h\[index\]<<1)){
int t=(index-lm\[nl\])\*(h\[index\]<<1);
printf("%d\\n",al\[lm\[nl\]\]\*2+t);
}else{
int t=(i-lm\[nl\])\*(h\[lm\[nl\]\]<<1);
printf("%d\\n",t+al\[lm\[nl\]\]);
}
}else if(h\[lm\[nl\]\]>h\[rm\[nr\]\]){
// debug("ok2\\n");
int i=nl;
bool fg=false;
int index=-1;
for(;h\[rm\[nr\]\]>=h\[i\];i--){
if(!fg&&h\[rm\[nr\]\]==h\[i\]){
fg=1;
index=i;
}
}
// debug("%d\\n",i);
if(fg&&ar\[rm\[nr\]\]<=(index-i)\*(h\[index\]<<1)){
int t=(rm\[nr\]-index)\*(h\[index\]<<1);
printf("%d\\n",ar\[rm\[nr\]\]\*2+t);
}else{
int t=(rm\[nr\]-i)\*(h\[rm\[nr\]\]<<1);
printf("%d\\n",t+ar\[rm\[nr\]\]);
}
}else{
int t=min(al\[lm\[nl\]\],ar\[rm\[nr\]\]);
// debug("%d %d %d\\n",al\[lm\[nl\]\],ar\[rm\[nr\]\],h\[rm\[nr\]\]);
printf("%d\\n",2\*t+(rm\[nr\]-lm\[nl\])\*(h\[rm\[nr\]\]<<1));
}
}
int main() {
ios\_base::sync\_with\_stdio(0);
int r,l;
while(~scanf("%d%d",&l,&r)&&r&&l){
n=(r-l)>>1;
for(int i=0,j=l;i<=n;i++,j+=2){
scanf("%d",&h\[i\]);
if(j==-1)nl=i;
if(j==1)nr=i;
}
solve();
// Set(ar,0);
// Set(al,0);
}
}