AtCoder Grand Contest 003 E - Sequential operations on Sequence
题目传送门:https://agc003.contest.atcoder.jp/tasks/agc003_e
题目大意
一串数,初始为\(1\sim N\),现有\(Q\)个操作,每次操作会把数组长度变成\(L_i\),多余的长度直接截断;长度不够则循环填充,问最后\(1\sim N\)每个数的出现次数
首先维护一个单调递增的栈,因为较短的\(L_i\)可以让较长的\(L_{i'}\)失去其意义
然后我们倒推,对于一个\(L_i\),它能对\(L_{i-1}\)产生\(F_i×\lfloor\dfrac{L_i}{L_{i-1}}\rfloor\)的贡献,那么剩余的\(L_i\%L_{i-1}\),我们可以递归处理,找到最大的小于其的\(L_j\),按同样的方法处理即可
记得使用差分
/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
static char buf[1000000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
int x=0,f=1; char ch=gc();
for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<3)+(x<<1)+ch-'0';
return x*f;
}
inline int read(){
int x=0,f=1; char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<3)+(x<<1)+ch-'0';
return x*f;
}
inline void print(int x){
if (x<0) putchar('-'),x=-x;
if (x>9) print(x/10);
putchar(x%10+'0');
}
const int N=1e5;
ll stack[N+10],f[N+10],delta[N+10];
int n,m,top;
void solve(ll x,ll v){
int tmp=upper_bound(stack+1,stack+1+top,x)-stack-1;
if (!tmp){
delta[1 ]+=v;
delta[x+1]-=v;
}else f[tmp]+=v*(x/stack[tmp]),solve(x%stack[tmp],v);
}
int main(){
n=read(),m=read();
stack[++top]=n;
for (int i=1;i<=m;i++){
ll x; scanf("%lld",&x);
while (top&&x<=stack[top]) top--;
stack[++top]=x;
}f[top]=1;
for (int i=top;i>1;i--) f[i-1]+=f[i]*(stack[i]/stack[i-1]),solve(stack[i]%stack[i-1],f[i]);
delta[1]+=f[1],delta[stack[1]+1]-=f[1];
for (int i=1;i<=n;i++) printf("%lld\n",delta[i]+=delta[i-1]);
return 0;
}