/* 标称并没有用到题解中提到的那种奇妙的性质,我们可以证明,正常从1开始走的话,需要T次,如何使这个次数减小?题解中提到一个办法,有一步小于n/t,我们考虑这一步,如果把它匀到左右两步中,则可以减小,就根据这个性质来优化 next函数的部分,我当时用了一个倍增法,题解用了一个并查集,倍增比较直观,然而并查集更为巧妙 */ //my code #include<iostream> #include<cstdio> #include<string> #include<cstring> #include<algorithm> #include<vector> #include<queue> #define ll long long using namespace std; const int maxn = 600500; int n,q,cnt,ct[maxn]; ll a[maxn],sum[maxn],ans; ll read(){ char ch=getchar(); ll x=0,f=1; while(!(ch>=‘0‘&&ch<=‘9‘)){if(ch==‘-‘)f=-1;ch=getchar();}; while(ch>=‘0‘&&ch<=‘9‘){x=x*10+(ch-‘0‘);ch=getchar();}; return x*f; } ll greedy(int u,ll t){ ll tot = 0,tmp = 0; ll j,lg,tst,nt; for(int i = 1;i <= n;i++){ tmp++; if(a[u] + a[u+1] > t || i == n){ }else{ lg = 0; j = 1; tot = 0; while(lg>=0){ if(u+j<=n){ tst = sum[u+j]-sum[u-1]; nt = u + j; } else{ tst = sum[n] - sum[u-1] + sum[u+j-n]; nt = u + j - n; } if(tot + tst <= t){ i += j; tot += tst - a[nt]; u = nt; lg++; j <<= 1; }else{ lg--; j >>= 1; } } } u++; if(u > n) u = 1; if(tmp >= ans) break; } return tmp; } void work(ll t){ cnt = 0; ans = maxn; ct[++cnt] = 1; ll sum = a[1]; for(int i = n;i > 1;i--){ sum += a[i]; if(sum > t) break; ct[++cnt] = i; } for(int i = 1;i <= cnt;i++){ ans = min(ans,greedy(ct[i],t)); } cout<<ans<<endl; } int main(){ freopen("stall.in","r",stdin); freopen("stall.out","w",stdout); n = read(); q = read(); for(int i = 1;i <= n;i++){ a[i] = read(); sum[i] = sum[i-1] + a[i]; } ll b; for(int i = 1;i <= q;i++){ b = read(); work(b); } return 0; } //std #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<vector> #include<queue> #include<map> #include<set> #include<stack> #include<cstdlib> #include<string> #include<bitset> #define INF 1000000000 #define N 2000005 #define fi first #define se second #define debug(x) cout<<#x<<"="<<x<<endl #define MP(x,y) make_pair(x,y) using namespace std; typedef long long LL; typedef pair<int,int> pii; LL a[N],s[N]; int to[N],fa[N],d[N]; int findr(int x) { if(fa[x]==x) return x; else { int t=fa[x]; fa[x]=findr(fa[x]); d[x]=d[x]+d[t]; return fa[x]; } } void Union(int x,int y) { fa[x]=y; d[x]=1; } int main() { int size = 256 << 20; // 256MB char *p = (char*)malloc(size) + size; __asm__("movl %0, %%esp\n" :: "r"(p)); freopen("stall.in","r",stdin); freopen("stall.out","w",stdout); cin>>n>>q; for(i=1;i<=n;i++) scanf("%I64d",&a[i]),a[i+n]=a[i]; for(i=n*2;i;i--) s[i]=s[i+1]+a[i]; while(q--) { ans=INF; cin>>m; memset(d,0,sizeof(d)); for(i=1;i<=n;i++) fa[i]=i,fa[i+n]=i+n; for(j=n*2,i=n*2;i;i--) { while(s[i]-s[j+1]>m) j--; to[i]=j+1; //debug(to[i]); } //if(q==1) return 0; for(i=1;i<=n;i++) { j=findr(i); while(to[j]<i+n) { k=to[j]; Union(j,k); j=findr(k); } findr(i); ans=min(ans,d[i]); } cout<<ans+1<<endl; } return 0; } // davidlee1999WTK 2015/ // srO myk Orz //ios::sync_with_stdio(false);
时间: 2024-10-21 14:24:37