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#184624#4322. rsraogps / 雪に咲く花LynkcatAC ✓0ms0kbC++172.5kb2023-09-20 23:53:592023-09-20 23:53:59

Judging History

你现在查看的是测评时间为 2023-09-20 23:53:59 的历史记录

  • [2023-09-20 23:58:59]
  • 管理员手动重测该提交记录
  • 测评结果:AC
  • 用时:614ms
  • 内存:230976kb
  • [2023-09-20 23:53:59]
  • 评测
  • 测评结果:0
  • 用时:0ms
  • 内存:0kb
  • [2023-09-20 23:53:59]
  • 提交

answer

#include<bits/stdc++.h>
#define poly vector<int>
#define IOS ios::sync_with_stdio(false)
#define ll long long 
#define mp make_pair
#define mt make_tuple
#define pa pair < unsigned int,unsigned int >
#define fi first
#define se second
#define inf 1e18
#define mod 998244353
#define N 1000005
using namespace std;
char obuf[1 << 27], *o(obuf);
inline char gc(){static char buf[1<<23],*p1=buf,*p2=buf;return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<23,stdin),p1==p2)?EOF:*p1++;}
// #define gc getchar
inline ll read(){char c=gc();ll su=0,f=1;for (;c<'0'||c>'9';c=gc()) if (c=='-') f=-1;for (;c>='0'&&c<='9';c=gc()) su=su*10+c-'0';return su*f;}
inline void write(ll x){if (x<0){putchar('-');write(-x);return;}if (x>=10) write(x/10);putchar(x%10+'0');}
inline void writesp(ll x){write(x),putchar(' ');}
inline void writeln(ll x){write(x);putchar('\n');}
void print(uint x) {
    if (x > 9)
        print(x / 10);
    *o++ = x % 10 + '0';
}
unsigned int n,m,a[N],b[N],c[N],ans[N*5];
unsigned int prea[N],preb[N],sum[N],sb[N];
pair<pa,int> all[N*5],all1[N*5];
inline int gcd(int x,int y)
{
	return (y==0)?x:gcd(y,x%y);
}
void BellaKira()
{
	n=read(),m=read();
	for (int i=1;i<=n;i++) a[i]=read();
	for (int i=1;i<=n;i++) b[i]=read();
	for (int i=1;i<=n;i++) c[i]=read();
	// m=2;
	for (int i=1;i<=m;i++)
	{
		int l=read(),r=read();
		sum[r]++;
		all[i]=mp(mp(r,l),i);
		all1[i]=all[i];
	}
	for (int i=1;i<=n;i++) 
		sum[i]+=sum[i-1];
	for (int i=1;i<=m;i++) 
		all[sum[all1[i].fi.fi]--]=all1[i];
	poly g;
	for (int i=1,pos=1;i<=n;i++)
	{
		int t=i;
		for (int j=i-1;j>=1;j--)
		{
			int x=a[j]&a[i];
			if (x==a[j]) break;
			if (j<t) sb[j]+=(prea[j]-prea[j-1])*(i-1),t=j;
			a[j]=x;
		}
		for (int j=i-1;j>=1;j--)
		{
			int x=b[j]|b[i];
			if (x==b[j]) break;
			if (j<t) sb[j]+=(prea[j]-prea[j-1])*(i-1),t=j;
			b[j]=x;
		}
		int lst=-1;
		for (int j=i-1;j>=1;j--)
		{
			if (lst!=c[j]&&c[i]%c[j]==0) break;
			lst=c[j];
			if (j<t) sb[j]+=(prea[j]-prea[j-1])*(i-1),t=j;
			c[j]=gcd(c[j],c[i]);
		}
		for (int j=t;j<=i;j++)
		{
			int v=a[j]*b[j]*c[j];
			sb[j]+=-v*(i-1);
			prea[j]=prea[j-1]+v,
			preb[j]=preb[j-1]+sb[j];
		}
		while (pos<=m&&all[pos].fi.fi==i)
		{
			int l=all[pos].fi.se,id=all[pos].se;
			{
				int A=prea[i]-prea[l-1];
				int B=preb[i]-preb[l-1];
				ans[id]+=A*i+B;
			}
			pos++;
		}
	}
	for (int i=1;i<=m;i++) print(ans[i]),
	*o++ = '\n';

    fwrite(obuf, o - obuf, 1, stdout);
}
signed main()
{
	int T=1;
	while (T--)
	{
		BellaKira();
	}
}

詳細信息

Test #1:

score: 0
Time Limit Exceeded

input:

This is the input of the first test case.

output:


result: