HDu 4578 Transformationセグメントツリー

6514 ワード

Transformation
Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 65535/65536 K (Java/Others) Total Submission(s): 3579 Accepted Submission(s): 859
Problem Description
Yuanfang is puzzled with the question below:
There are n integers, a
1, a
2, …, a
n. The initial values of them are 0. There are four kinds of operations.
Operation 1: Add c to each number between a
x and a
y inclusive. In other words, do transformation a
k<---a
k+c, k = x,x+1,…,y.
Operation 2: Multiply c to each number between a
x and a
y inclusive. In other words, do transformation a
k<---a
k×c, k = x,x+1,…,y.
Operation 3: Change the numbers between a
x and a
y to c, inclusive. In other words, do transformation a
k<---c, k = x,x+1,…,y.
Operation 4: Get the sum of p power among the numbers between a
x and a
y inclusive. In other words, get the result of a
x
p+a
x+1
p+…+a
y
p.
Yuanfang has no idea of how to do it. So he wants to ask you to help him.

Input
There are no more than 10 test cases.
For each case, the first line contains two numbers n and m, meaning that there are n integers and m operations. 1 <= n, m <= 100,000.
Each the following m lines contains an operation. Operation 1 to 3 is in this format: "1 x y c"or "2 x y c"or "3 x y c". Operation 4 is in this format: "4 x y p". (1 <= x <= y <= n, 1 <= c <= 10,000, 1 <= p <= 3)
The input ends with 0 0.

Output
For each operation 4, output a single integer in one line representing the result. The answer may be quite large. You just need to calculate the remainder of the answer when divided by 10007.

Sample Input

Sample Output

Source
2013 ACM-ICPC杭州試合区全国招待試合
セグメントツリー:set,mul,addの3つのタグがあります.
setの場合、mul,addを初期化します.
mul操作の場合、mul*mul、add*mulを直接マージできます.それから各区間はまず乗算をして、それから加算をします

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 300000
#define ll long long
int sum[maxn][3];
int mul[maxn],add[maxn],val[maxn];
int lc[maxn],rc[maxn],set[maxn];
int mod = 10007;
int cnt;
void init(){
    cnt = 1;
    memset(sum,0,sizeof(sum));
    memset(lc,0,sizeof(lc));
    memset(rc,0,sizeof(rc));
    memset(val,0,sizeof(val));
    for(int i = 0;i < maxn; i++)
        mul[i] = 1;
    memset(add,0,sizeof(add));
    memset(set,0,sizeof(set));
}

void update(int u,int l,int r){
    if(l == r || u == 0) return ;
    int len = (r-l+1)%mod;
    if(set[u] != 0){
        int num = set[u]*set[u]%mod;
        sum[u][0] = set[u]*len%mod;
        sum[u][1] = num*len%mod;
        sum[u][2] = num*set[u]%mod*len%mod;
        return ;
    }
    int s1 = (sum[lc[u]][0] + sum[rc[u]][0])%mod;
    int s2 = (sum[lc[u]][1] + sum[rc[u]][1])%mod;
    int s3 = (sum[lc[u]][2] + sum[rc[u]][2])%mod;
    if(mul[u] > 1){
        s1 = (ll)s1*mul[u]%mod;
        s2 = (ll)s2*mul[u]*mul[u]%mod;
        s3 = (ll)s3*mul[u]*mul[u]*mul[u]%mod;
    }
    if(add[u] != 0){
        ll a1 = add[u],a2= a1*a1,a3=a2*a1;
        sum[u][0] = (ll)(s1 + a1*len)%mod;
        sum[u][1] = (ll)(s2+2*s1*a1+a2*len)%mod;
        sum[u][2] = ((ll)s3+3*s2*a1+3*s1*a2+a3*len)%mod;
    }
    else sum[u][0] = s1,sum[u][1] = s2,sum[u][2] = s3;
}
void pushdown(int u,int l,int r,int op,int num){
    if(u == 0) return ;
    if(l == r){
        if(op == 1) val[u] = (val[u] + num) % mod;
        else if(op == 2) val[u] = val[u] * num % mod;
        else if(op == 3) val[u] = num;
        sum[u][0] = val[u];
        sum[u][1] = (ll)val[u]*val[u]%mod;
        sum[u][2] = (ll)val[u]*val[u]*val[u]%mod;
        return ;
    }
    if(op == 3){
        set[u] = num;
        add[u] = 0;
        mul[u] = 1;
        update(u,l,r);
        return ;
    }
    int mid = (l+r)/2;
    if(set[u] != 0){
        pushdown(lc[u],l,mid,3,set[u]);
        pushdown(rc[u],mid+1,r,3,set[u]);
        set[u] = 0;
    }
    if(op == 2){
        mul[u] = mul[u]*num%mod;
        add[u] = add[u]*num%mod;
    }
    if(op == 1){
        if(num == 0){
            if(mul[u] != 1){
                pushdown(lc[u],l,mid,2,mul[u]);
                pushdown(rc[u],mid+1,r,2,mul[u]);
                mul[u] = 1;
            }
            if(add[u] > 0){
                pushdown(lc[u],l,mid,1,add[u]);
                pushdown(rc[u],mid+1,r,1,add[u]);
            }
            add[u] = 0;
        }
        else  add[u] = (add[u]+num)%mod;
    }
    update(u,l,r);
}
void build(int u,int l,int r){
    if(l == r) return ;
    int mid = (l+r)/2;
    lc[u] = cnt++;
    rc[u] = cnt++;
    build(lc[u],l,mid);
    build(rc[u],mid+1,r);
}

int query(int u,int l,int r,int L,int R,int ty){
    if(l == L && R == r)
        return sum[u][ty];
    pushdown(u,l,r,1,0);
    int mid = (l+r)/2;
    int ans = 0;
    if(mid < L) ans = query(rc[u],mid+1,r,L,R,ty);
    else if(mid >= R) ans = query(lc[u],l,mid,L,R,ty);
    else ans = query(lc[u],l,mid,L,mid,ty)+query(rc[u],mid+1,r,mid+1,R,ty);
    update(u,l,r);
    return ans%mod;
}

void work(int u,int l,int r,int L,int R,int op,int num){
    if(l == L && R == r){
        pushdown(u,l,r,op,num);
        return ;
    }
    pushdown(u,l,r,1,0);
    int mid = (l+r)/2;
    if(mid < L) work(rc[u],mid+1,r,L,R,op,num);
    else if(mid >= R) work(lc[u],l,mid,L,R,op,num);
    else work(lc[u],l,mid,L,mid,op,num),work(rc[u],mid+1,r,mid+1,R,op,num);
    update(u,l,r);
}


int main(){
    int n,m;
    //freopen("4578.in","r",stdin);
    //freopen("4578.out","w",stdout);
    while(scanf("%d%d",&n,&m),n+m){
        int op,l,r,num;
        init();
        cnt++;
        build(1,1,n);
        while(m--){
            scanf("%d%d%d%d",&op,&l,&r,&num);
            if(op != 4)  work(1,1,n,l,r,op,num%mod);
            else printf("%d
",query(1,1,n,l,r,num-1));
        }
    }
    return 0;
}

JAvaシーケンス化——Serializable

ローカルメソッド呼び出し3