HDU - 4578 Transformation（線段樹區間修改）
阿新 • • 發佈：2018-09-02
大神 題意 span upd https blank ring 表示 while
https://cn.vjudge.net/problem/HDU-4578
題意
4種操作，區間加，區間乘，區間變為一個數，求區間的和、平方和以及立方和。
分析
明顯線段樹，不過很麻煩。。看kuangbin大神的代碼打的
用sum1,sum2,sum3分別代表和、平方和、立方和。
懶惰標記使用三個變量：
lazy1:是加的數
lazy2:是乘的倍數
lazy3:是賦值為一個常數，為0表示沒有。
#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include<vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> using namespace std; const int MOD = 10007; const int MAXN = 100010; struct Node { int l,r; int sum1,sum2,sum3; int lazy1,lazy2,lazy3; } segTree[MAXN*3]; void build(int i,int l,int r) { segTree[i].l = l; segTree[i].r = r; segTree[i].sum1 = segTree[i].sum2 = segTree[i].sum3 = 0; segTree[i].lazy1 = segTree[i].lazy3 = 0; segTree[i].lazy2 = 1; int mid = (l+r)/2; if(l == r)return; build(i<<1,l,mid); build((i<<1)|1,mid+1,r); } void push_up(int i) { if(segTree[i].l == segTree[i].r) return; segTree[i].sum1 = (segTree[i<<1].sum1 + segTree[(i<<1)|1].sum1)%MOD; segTree[i].sum2 = (segTree[i<<1].sum2 + segTree[(i<<1)|1].sum2)%MOD; segTree[i].sum3 = (segTree[i<<1].sum3 + segTree[(i<<1)|1].sum3)%MOD; } void push_down(int i) { if(segTree[i].l == segTree[i].r) return; if(segTree[i].lazy3 != 0) { segTree[i<<1].lazy3 = segTree[(i<<1)|1].lazy3 = segTree[i].lazy3; segTree[i<<1].lazy1 = segTree[(i<<1)|1].lazy1 = 0; segTree[i<<1].lazy2 = segTree[(i<<1)|1].lazy2 = 1; segTree[i<<1].sum1 = (segTree[i<<1].r - segTree[i<<1].l + 1)*segTree[i<<1].lazy3%MOD; segTree[i<<1].sum2 = (segTree[i<<1].r - segTree[i<<1].l + 1)*segTree[i<<1].lazy3%MOD*segTree[i<<1].lazy3%MOD; segTree[i<<1].sum3 = (segTree[i<<1].r - segTree[i<<1].l + 1)*segTree[i<<1].lazy3%MOD*segTree[i<<1].lazy3%MOD*segTree[i<<1].lazy3%MOD; segTree[(i<<1)|1].sum1 = (segTree[(i<<1)|1].r - segTree[(i<<1)|1].l + 1)*segTree[(i<<1)|1].lazy3%MOD; segTree[(i<<1)|1].sum2 = (segTree[(i<<1)|1].r - segTree[(i<<1)|1].l + 1)*segTree[(i<<1)|1].lazy3%MOD*segTree[(i<<1)|1].lazy3%MOD; segTree[(i<<1)|1].sum3 = (segTree[(i<<1)|1].r - segTree[(i<<1)|1].l + 1)*segTree[(i<<1)|1].lazy3%MOD*segTree[(i<<1)|1].lazy3%MOD*segTree[(i<<1)|1].lazy3%MOD; segTree[i].lazy3 = 0; } if(segTree[i].lazy1 != 0 || segTree[i].lazy2 != 1) { segTree[i<<1].lazy1 = ( segTree[i].lazy2*segTree[i<<1].lazy1%MOD + segTree[i].lazy1 )%MOD; segTree[i<<1].lazy2 = segTree[i<<1].lazy2*segTree[i].lazy2%MOD; int sum1,sum2,sum3; sum1 = (segTree[i<<1].sum1*segTree[i].lazy2%MOD + (segTree[i<<1].r - segTree[i<<1].l + 1)*segTree[i].lazy1%MOD)%MOD; sum2 = (segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[i<<1].sum2 % MOD + 2*segTree[i].lazy1*segTree[i].lazy2%MOD * segTree[i<<1].sum1%MOD + (segTree[i<<1].r - segTree[i<<1].l + 1)*segTree[i].lazy1%MOD*segTree[i].lazy1%MOD)%MOD; sum3 = segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[i<<1].sum3 % MOD; sum3 = (sum3 + 3*segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[i<<1].sum2) % MOD; sum3 = (sum3 + 3*segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD * segTree[i<<1].sum1) % MOD; sum3 = (sum3 + (segTree[i<<1].r - segTree[i<<1].l + 1)*segTree[i].lazy1%MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD) % MOD; segTree[i<<1].sum1 = sum1; segTree[i<<1].sum2 = sum2; segTree[i<<1].sum3 = sum3; segTree[i<<1|1].lazy1 = ( segTree[i].lazy2*segTree[i<<1|1].lazy1%MOD + segTree[i].lazy1 )%MOD; segTree[i<<1|1].lazy2 = segTree[i<<1|1].lazy2*segTree[i].lazy2%MOD; sum1 = (segTree[i<<1|1].sum1*segTree[i].lazy2%MOD + (segTree[i<<1|1].r - segTree[i<<1|1].l + 1)*segTree[i].lazy1%MOD)%MOD; sum2 = (segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[i<<1|1].sum2 % MOD + 2*segTree[i].lazy1*segTree[i].lazy2%MOD * segTree[i<<1|1].sum1%MOD + (segTree[i<<1|1].r - segTree[i<<1|1].l + 1)*segTree[i].lazy1%MOD*segTree[i].lazy1%MOD)%MOD; sum3 = segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[i<<1|1].sum3 % MOD; sum3 = (sum3 + 3*segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[i<<1|1].sum2) % MOD; sum3 = (sum3 + 3*segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD * segTree[i<<1|1].sum1) % MOD; sum3 = (sum3 + (segTree[i<<1|1].r - segTree[i<<1|1].l + 1)*segTree[i].lazy1%MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD) % MOD; segTree[i<<1|1].sum1 = sum1; segTree[i<<1|1].sum2 = sum2; segTree[i<<1|1].sum3 = sum3; segTree[i].lazy1 = 0; segTree[i].lazy2 = 1; } } void update(int i,int l,int r,int type,int c) { if(segTree[i].l >= l && segTree[i].r <= r) { c %= MOD; if(type == 1) { segTree[i].lazy1 += c; segTree[i].lazy1 %= MOD; segTree[i].sum3 = (segTree[i].sum3 + 3*segTree[i].sum2%MOD*c%MOD + 3*segTree[i].sum1%MOD*c%MOD*c%MOD + (segTree[i].r - segTree[i].l + 1)*c%MOD*c%MOD*c%MOD)%MOD; segTree[i].sum2 = (segTree[i].sum2 + 2*segTree[i].sum1%MOD*c%MOD + (segTree[i].r - segTree[i].l + 1)*c%MOD*c%MOD)%MOD; segTree[i].sum1 = (segTree[i].sum1 + (segTree[i].r - segTree[i].l + 1)*c%MOD)%MOD; } else if(type == 2) { segTree[i].lazy1 = segTree[i].lazy1*c%MOD; segTree[i].lazy2 = segTree[i].lazy2*c%MOD; segTree[i].sum1 = segTree[i].sum1*c%MOD; segTree[i].sum2 = segTree[i].sum2*c%MOD*c%MOD; segTree[i].sum3 = segTree[i].sum3*c%MOD*c%MOD*c%MOD; } else { segTree[i].lazy1 = 0; segTree[i].lazy2 = 1; segTree[i].lazy3 = c%MOD; segTree[i].sum1 = c*(segTree[i].r - segTree[i].l + 1)%MOD; segTree[i].sum2 = c*(segTree[i].r - segTree[i].l + 1)%MOD*c%MOD; segTree[i].sum3 = c*(segTree[i].r - segTree[i].l + 1)%MOD*c%MOD*c%MOD; } return; } push_down(i); int mid = (segTree[i].l + segTree[i].r)/2; if(l <= mid)update(i<<1,l,r,type,c); if(r > mid)update((i<<1)|1,l,r,type,c); push_up(i); } int query(int i,int l,int r,int p) { if(segTree[i].l >= l && segTree[i].r <= r) { if(p == 1)return segTree[i].sum1; else if(p== 2)return segTree[i].sum2; else return segTree[i].sum3; } push_down(i); int mid = (segTree[i].l + segTree[i].r )/2; int sum=0; if(l <= mid) sum+=query(i<<1,l,r,p); if(r > mid) sum+=query((i<<1)|1,l,r,p); return sum%MOD; } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int n,m; while(scanf("%d%d",&n,&m) == 2) { if(n == 0 && m == 0)break; build(1,1,n); int type,x,y,c; while(m--) { scanf("%d%d%d%d",&type,&x,&y,&c); if(type == 4)printf("%d\n",query(1,x,y,c)); else update(1,x,y,type,c); } } return 0; }
HDU - 4578 Transformation（線段樹區間修改）