Fourth multi-school Competition

P1006: a line segment tree without lazy.

The idea is latency mark.

#include<stdio.h>#include<algorithm>#include<string.h>#include<math.h>#include<iostream>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define N 111111typedef long long ll;ll a[N*4],add[N*4];int n;int gcd(int a,int b){    if (a%b==0)  return b;    return gcd(b,a%b);}void build(int l,int r,int rt){    add[rt]=0;    if (l==r)    {        scanf("%d",&a[rt]);        return;    }    int m=(l+r)>>1;    build(lson);    build(rson);}void pushdown(int rt){    if (add[rt])    {        add[rt<<1]=add[rt<<1|1]=add[rt];        a[rt<<1]=a[rt<<1|1]=a[rt];        add[rt]=0;    }}void update1(int L,int R,int l,int r,int rt,int x){    if (L<=l&&R>=r)    {        add[rt]=1;        a[rt]=x;        return;    }    pushdown(rt);    int m=(l+r)>>1;    if (L<=m) update1(L,R,lson,x);    if (R>m)  update1(L,R,rson,x);  }void update2(int L,int R,int l,int r,int rt,int x){   if (add[rt]&&L<=l&&R>=r)   {           if (a[rt]>x)           a[rt]=gcd(a[rt],x);           return;   }   if (l==r)   {       if (a[rt]>x) a[rt]=gcd(a[rt],x);       return ;   }   pushdown(rt);   int m=(l+r)>>1;   if (L<=m) update2(L,R,lson,x);   if (R>m) update2(L,R,rson,x);}int query(int x,int l,int r,int rt){    if (l==r) return a[rt];    int m=(l+r)>>1;    pushdown(rt);    if (x<=m) return query(x,lson);    else    if (x>m)  return query(x,rson);}int main(){    int T;    scanf("%d",&T);    while (T--){    scanf("%d",&n);    build(1,n,1);           int Q;           scanf("%d",&Q);           while (Q--)           {               int t,l,r,x;               scanf("%d%d%d%d",&t,&l,&r,&x);               if (t==1) update1(l,r,1,n,1,x);                    else update2(l,r,1,n,1,x);            }         for (int i=1;i<=n;i++)         printf("%d ",query(i,1,n,1));         printf("\n");        }return 0;}

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P1005: Status DP has not been called,

It turns out that there is a mod operation in the equation, and then there is a negative when the subtraction occurs, Nima is going crazy.

# Include <stdio. h> # include <algorithm> # include <string. h> # include <math. h> typedef long ll; using namespace STD; # define mod limit 7ll L [1234] [1124], R [1234] [1124]; int A [1234]; int N; int main () {int t; scanf ("% d", & T); While (t --) {scanf ("% d", & N); memset (L, 0, sizeof (l); memset (R, 0, sizeof (R); For (INT I = 1; I <= N; I ++) scanf ("% d", & A [I]); // records the statuses on both sides of the For (INT I = N; I> = 1; I --) {R [I] [A [I] ++; For (Int J = 0; j <1024; j ++) if (R [I + 1] [J]) {R [I] [J] + = R [I + 1] [J]; if (R [I] [J]> mod) R [I] [J]-= MOD; R [I] [J & A [I] + = R [I + 1] [J]; if (R [I] [J & A [I]> mod) R [I] [J & A [I]-= MOD ;}} for (INT I = 1; I <= N; I ++) {L [I] [A [I] ++; For (Int J = 0; j <1024; j ++) if (L [I-1] [J]) {L [I] [J] + = L [I-1] [J]; if (L [I] [J]> mod) L [I] [J]-= MOD; L [I] [J ^ A [I] + = L [I-1] [J]; If (L [I] [J ^ A [I]> mod) L [I] [J ^ A [I]-= mod ;}} ll ans = 0; For (INT I = 1; I <n; I ++) for (Int J = 0; j <1024; j ++) {ll OK = L [I] [J]-l [I-1] [J]; // The most important step is to ensure that the first I has been selected, and then match the right with if (OK <0) OK + = MOD; If () OK = OK * R [I + 1] [J] % MOD; ans + = OK; If (ANS> mod) ans-= MOD ;} printf ("% i64d \ n", ANS );}}

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