CF1119E

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题意:你有长度为$2^0,2^1\dots 2^{n-1}$的木条分别$a_0,a_1,\dots a_{n-1}$根。

求最多能组成多少个三角形,使得每根木条最多用在一个三角形内,且每个三角形每条边仅有一根木条构成。$n\le 3\cdot 10^5,a_i\le 10^9$

发现每个三条边可以构成三角形当且仅当形如$2^x,2^y,2^y$其中$x\le y$。

猜一个贪心,从小到大遍历每个长度$2^i$,如果可以用两根和前面剩下的组一个就组,否则如果有三根一样的就组一个。

这样写能过,尝试证明:

假设有$2^a$3根并且不存在$\lt 2^a$的,证明不存在一种情况使得把它们拆开比不拆答案大

  1. 这三根只有0根在后面被拼成三角形,显然把他们拼起来答案会+1
  2. 这三根只有1根在后面被拼成三角形,显然把他们拼起来而放弃其中一根组成的三角形答案不会变小
  3. 这三根只有2根在后面被拼成三角形,设分别和$2^x,2^y$各两根组$(x\le y)$,那么把$2^x,2^y,2^y$组一组而把这三根拼起来答案是一样的。
  4. 这三根全部在后面被拼成三角形,设分别和$2^x,2^y,2^z$各两根组$(x\le y\le z)$,那么把$(2^x,2^y,2^y)(2^x,2^z,2^z)$组起来而把这三根拼起来答案是一样的。

所以这样是对的。

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/*
Author: QAQ-Automaton
LANG: C++
PROG: E.cpp
Mail: cnyalilk@vip.qq.com
*/
#include<bits/stdc++.h>
#define int long long
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define inf 0x3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
const int SIZE = (1 << 21) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
// getchar
#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
// print the remaining part
inline void flush () {
fwrite (obuf, 1, oS - obuf, stdout);
oS = obuf;
}
// putchar
inline void putc (char x) {
*oS ++ = x;
if (oS == oT) flush ();
}
// input a signed integer
inline bool read (signed &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}

inline bool read (long long &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}
inline bool read (char &x) {
x=gc();
return x!=EOF;
}
inline bool read(char *x){
while((*x=gc())=='\n' || *x==' '||*x=='\r')if(*x==EOF)return 0;
while(!(*x=='\n'||*x==' '||*x=='\r'))*(++x)=gc();
*x=0;
return 1;
}
template<typename A,typename ...B>
inline bool read(A &x,B &...y){
return read(x)&&read(y...);
}
// print a signed integer
inline bool write (signed x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}

inline bool write (long long x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}
inline bool write (char x) {
putc(x);
return 0;
}
inline bool write(const char *x){
while(*x){putc(*x);++x;}
return 0;
}
inline bool write(char *x){
while(*x){putc(*x);++x;}
return 0;
}
template<typename A,typename ...B>
inline bool write(A x,B ...y){
return write(x)||write(y...);
}
//no need to call flush at the end manually!
struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
signed main(){
#ifdef QAQAutoMaton
freopen("E.in","r",stdin);
freopen("E.out","w",stdout);
#endif
int n;
read(n);
int s=0,t=0,w,x;
for(int i=1;i<=n;++i){
read(w);
s+=x=min(t,w/2);
t-=x;
w-=x+x;
s+=w/3;
t+=w%3;
}
write(s,'\n');
return 0;
}

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