题解 -
题目意思
- 真没想到 卡卡常竟然过了,虽然有更优秀的解法。
考虑枚举第一段的起点 以及用 表示以其作为起点的最小分组。以及假设 为序列的最后部分与开始部分接起来的值,那么答案就考虑如果 即为 否则为 (相当于把最后一段再分一次)
对于每次更新 时候记录 为当前最小分组时候最远能达到的点, 具体求法可以看代码。
#pragma GCC optimize(3,"Ofast","inline")
#pragma GCC target("avx,avx2")
#include <bits/stdc++.h>
#define For(i,a,b) for ( register int i=(a);i<=(b);i++ )
#define Dow(i,b,a) for ( int i=(b);i>=(a);i-- )
#define GO(i,x) for ( int i=head[x];i;i=e[i].nex )
#define mem(x,s) memset(x,s,sizeof(x))
#define cpy(x,s) memcpy(x,s,sizeof(x))
#define YES return puts("YES"),0
#define NO return puts("NO"),0
#define GG return puts("-1"),0
#define pb push_back
using namespace std;
struct IO
{
#define MAXSIZE (1 << 20)
#define isdigit(x) (x >= '0' && x <= '9')
char buf[MAXSIZE],*p1,*p2;
char pbuf[MAXSIZE],*pp;
IO():p1(buf),p2(buf),pp(pbuf){}
inline char gc() {
// return getchar();
if (p1==p2) p2=(p1=buf)+fread(buf,1,MAXSIZE,stdin);
return p1==p2?' ':*p1++;
}
inline bool blank(char ch) {return ch==' '||ch =='\n'||ch == '\r'||ch == '\t';}
template <class T>
inline void read(T &x)
{
register double tmp=1;register bool sign=0; x=0;
register char ch=gc();
for (;!isdigit(ch);ch=gc()) if(ch=='-') sign=1;
for (;isdigit(ch);ch=gc()) x=x*10+(ch-'0');
if (ch=='.') for (ch=gc();isdigit(ch);ch=gc()) tmp/=10.0,x+=tmp*(ch-'0');
if (sign) x=-x;
}
inline void read(char *s)
{
register char ch=gc();
for (;blank(ch);ch=gc());
for (;!blank(ch);ch=gc()) *s++=ch;
*s=0;
}
inline void read(char &c) {for(c=gc();blank(c);c=gc());}
template<class t> inline void write(t x){
if(x<0) putchar('-'),write(-x);
else{if(x>9) write(x/10);putchar('0'+x%10);}
}
} io;
const int mod=1e9+7;
const int mo=998244353;
const int N=1e6+5;
int n,m,a[N],f[N],g[N];
long long b[N];
inline int min(int x,int y) { return (x>y)?y:x;}
signed main()
{
io.read(n),io.read(m);
For(i,1,n)
{
io.read(a[i]);
b[i]=b[i-1]+a[i];
}
for (;m--;)
{
long long x;
io.read(x);
if(b[n]<=x)
{
puts("1");
continue;
}
f[n+1]=0;
Dow(i,n,1)
{
g[i]=i;
int l=i,r=n;
f[i]=f[i+1]+1;
while(l<=r)
{
int mid=l+r>>1;
if(b[mid]-b[i-1]<=x) g[i]=mid,l=mid+1;
else r=mid-1;
}
f[i]=f[g[i]+1]+1;
if(g[i]==n) g[i]=i;
else if(g[i]+1<=n) g[i]=g[g[i]+1];
}
int ans=1e9;
For(i,1,n) if(b[i-1]<=x)
{
long long half=b[n]-b[g[i]-1];
if(half+b[i-1]<=x) ans=min(ans,f[i]);
else ans=min(ans,f[i]+1);
}
else break;
io.write(ans),putchar('\n');
}
return 0;
}