Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 65535/65536 K (Java/Others)
Total Submission(s): 3830 Accepted Submission(s): 940
#include<bits/stdc++.h>
using namespace std;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
typedef long long INT;
const int maxn = 120000;
const int mod = 10007;
struct SegTree{
int add,multi,setv;
int sum1,sum2,sum3;
}segs[maxn*4];
void PushUp(int rt,int L,int R){
segs[rt].sum1 = (segs[rt*2].sum1 + segs[rt*2+1].sum1)% mod;
segs[rt].sum2 = (segs[rt*2].sum2 + segs[rt*2+1].sum2)% mod;
segs[rt].sum3 = (segs[rt*2].sum3 + segs[rt*2+1].sum3)% mod;
}
void buildtree(int rt,int L,int R){
segs[rt].multi = 1;
segs[rt].add = segs[rt].setv = 0;
segs[rt].sum1 = segs[rt].sum2 = segs[rt].sum3 = 0;
if(L == R){
return ;
}
buildtree(lson);
buildtree(rson);
}
void PushDown(int rt,int L,int R){
if(segs[rt].setv){ //考虑下放赋值标记
segs[rt*2].setv = segs[rt*2+1].setv = segs[rt].setv;
segs[rt*2].add = segs[rt*2+1].add = 0;
segs[rt*2].multi = segs[rt*2+1].multi = 1;
segs[rt*2].sum1 = (mid-L+1)*segs[rt].setv % mod;
segs[rt*2].sum2 = (mid-L+1)*segs[rt].setv %mod *segs[rt].setv % mod;
segs[rt*2].sum3 = (mid-L+1)*segs[rt].setv %mod *segs[rt].setv % mod *segs[rt].setv % mod;
segs[rt*2+1].sum1 = (R-mid)*segs[rt].setv % mod;
segs[rt*2+1].sum2 = (R-mid)*segs[rt].setv % mod *segs[rt].setv % mod;
segs[rt*2+1].sum3 = (R-mid)*segs[rt].setv % mod *segs[rt].setv % mod *segs[rt].setv % mod;
segs[rt].setv = 0;
}
if(segs[rt].multi != 1 || segs[rt].add){//如果有加和标记或者乘积标记
segs[rt*2].add = (segs[rt].multi * segs[rt*2].add %mod + segs[rt].add) % mod;
segs[rt*2].multi = segs[rt].multi * segs[rt*2].multi % mod;
int sum1, sum2 ,sum3;
//一次方和
sum1 = (segs[rt*2].sum1*segs[rt].multi %mod + (mid-L+1)*segs[rt].add %mod) % mod;
//平方和
sum2 = (segs[rt*2].sum2*segs[rt].multi %mod *segs[rt].multi %mod + 2*segs[rt].add * segs[rt].multi %mod *segs[rt*2].sum1 %mod + (mid-L+1)*segs[rt].add %mod *segs[rt].add %mod ) % mod;
//三次方和
sum3 = segs[rt*2].sum3*segs[rt].multi %mod *segs[rt].multi %mod *segs[rt].multi %mod;
sum3 = (sum3 + 3*segs[rt].add*segs[rt].multi %mod *segs[rt].multi %mod *segs[rt*2].sum2 %mod) % mod;
sum3 = (sum3 + 3*segs[rt].add*segs[rt].add %mod *segs[rt].multi %mod *segs[rt*2].sum1 %mod ) % mod;
sum3 = (sum3 + (mid-L+1)*segs[rt].add %mod *segs[rt].add %mod *segs[rt].add %mod ) % mod;
segs[rt*2].sum1 = sum1;
segs[rt*2].sum2 = sum2;
segs[rt*2].sum3 = sum3;
//同理,更新右儿子
segs[rt*2+1].add = (segs[rt*2+1].add*segs[rt].multi %mod + segs[rt].add) % mod;
segs[rt*2+1].multi = segs[rt*2+1].multi * segs[rt].multi % mod;
sum1 = (segs[rt*2+1].sum1*segs[rt].multi %mod + (R-mid)*segs[rt].add %mod) % mod;
sum2 = (segs[rt*2+1].sum2*segs[rt].multi %mod *segs[rt].multi %mod + 2*segs[rt].add*segs[rt].multi %mod *segs[rt*2+1].sum1 %mod + (R-mid)*segs[rt].add %mod *segs[rt].add %mod) % mod;
sum3 = segs[rt*2+1].sum3*segs[rt].multi %mod *segs[rt].multi %mod *segs[rt].multi %mod;
sum3 = (sum3 + 3*segs[rt].add*segs[rt].multi %mod *segs[rt].multi %mod *segs[rt*2+1].sum2 %mod) % mod;
sum3 = (sum3 + 3*segs[rt].add*segs[rt].add %mod *segs[rt].multi %mod *segs[rt*2+1].sum1 %mod ) % mod;
sum3 = (sum3 + (R-mid)*segs[rt].add %mod *segs[rt].add %mod *segs[rt].add %mod ) % mod;
segs[rt*2+1].sum1 = sum1;
segs[rt*2+1].sum2 = sum2;
segs[rt*2+1].sum3 = sum3;
segs[rt].add = 0;
segs[rt].multi = 1;
}
}
void Update(int rt,int L,int R,int l_ran,int r_ran,int c,int typ){
if(l_ran <= L&&R <= r_ran){
if(typ == 1){ //加和
segs[rt].add = (segs[rt].add + c)%mod;
segs[rt].sum3 = (segs[rt].sum3 + (R-L+1)*c %mod *c %mod *c %mod + 3*c %mod *segs[rt].sum2 %mod + 3*c %mod *c %mod *segs[rt].sum1 %mod ) % mod;
segs[rt].sum2 = (segs[rt].sum2 + (R-L+1)*c %mod *c %mod + 2*segs[rt].sum1 %mod *c %mod ) % mod;
segs[rt].sum1 = ((R-L+1)*c %mod + segs[rt].sum1) % mod;
}else if(typ == 2){ //乘积
//乘积对当前节点的和跟乘都有影响
segs[rt].add = segs[rt].add * c % mod;
segs[rt].multi = segs[rt].multi * c % mod;
segs[rt].sum1 = segs[rt].sum1 * c % mod;
segs[rt].sum2 = segs[rt].sum2 * c %mod *c %mod;
segs[rt].sum3 = segs[rt].sum3 *c %mod *c %mod *c % mod;
}else{ //赋值
segs[rt].setv = c;
segs[rt].multi = 1;
segs[rt].add = 0;
segs[rt].sum1 = c*(R-L+1) % mod;
segs[rt].sum2 = c*(R-L+1) % mod*c % mod;
segs[rt].sum3 = c*(R-L+1) % mod*c % mod *c %mod ;
}
return ;
}
PushDown(rt,L,R);
if(l_ran <= mid)
Update(lson,l_ran,r_ran,c,typ);
if(r_ran > mid)
Update(rson,l_ran,r_ran,c,typ);
PushUp(rt,L,R);
}
int query(int rt,int L,int R,int l_ran,int r_ran,int pw){
if(l_ran <= L&&R <= r_ran){
if(pw == 1){
return segs[rt].sum1;
}else if(pw == 2){
return segs[rt].sum2;
}else{
return segs[rt].sum3;
}
}
PushDown(rt,L,R);
int ret = 0;
if(l_ran <= mid){
ret = (ret + query(lson,l_ran,r_ran,pw)) % mod;
}
if(r_ran > mid){
ret = (ret + query(rson,l_ran,r_ran,pw)) % mod;
}
return ret;
}
int main(){
int n,m;
while(scanf("%d%d",&n,&m)!=EOF&&(n+m)){
int c,u,v,w;
buildtree(1,1,n);
for(int i = 1; i <= m; i++){
scanf("%d%d%d%d",&c,&u,&v,&w);
if(c == 4){
int res = query(1,1,n,u,v,w);
printf("%d\n",res);
}else {
Update(1,1,n,u,v,w,c);
}
}
}
return 0;
}
HDU 4578——Transformation——————【线段树区间操作、确定操作顺序】
原文:http://www.cnblogs.com/chengsheng/p/4975396.html