[线段树+区间求和]LightOJ 1183 - Computing Fast Average

题目链接：http://lightoj.com/volume_showproblem.php?problem=1183

1183 - Computing Fast Average

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Time Limit:2 second(s)

Memory Limit:64 MB

Given an array of integers (0indexed), you have to perform two types of queries in the array.

1.1 i j v- change the value of the elements fromi^thindex toj^thindex tov.

2.2 i j- find the average value of the integers fromi^thindex toj^thindex.

You can assume that initially all the values in the array are0.

Input

Input starts with an integerT (≤ 5), denoting the number of test cases.

Each case contains two integers:n (1 ≤ n ≤ 10⁵), q (1 ≤ q ≤ 50000), wherendenotes the size of the array. Each of the nextqlines will contain a query of the form:

1 i j v (0 ≤ i ≤ j < n, 0 ≤ v ≤ 10000)

2 i j (0 ≤ i ≤ j < n)

Output

For each case, print the case number first. Then for each query of the form'2 i j'print the average value of the integers fromitoj. If the result is an integer, print it. Otherwise print the result in'x/y'form, wherexdenotes the numerator andydenotes the denominator of the result andxandyare relatively prime.

Sample Input	Output for Sample Input
1 10 6 1 0 6 6 2 0 1 1 1 1 2 2 0 5 1 0 3 7 2 0 1	Case 1: 6 16/3 7

Note

Dataset is huge. Use faster i/o methods.

题意：给定一个序列，要求可以成段更新和求区间的平均数。

分析：稍微做了点变化，要求输出最简分式，用gcd约下分就可以了。

线段树：点树-成段更新-区间询问。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 111111;
int sum[MAXN<<2],cov[MAXN<<2],t,n,q;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
void build(){
    memset(sum,0,sizeof(sum));
    memset(cov,-1,sizeof(cov));
}
void pushDOWN(int rt,int l,int r){
    if(cov[rt]!=-1){
        int m = (l+r)>>1;
        sum[rt<<1] = (m-l+1)*cov[rt];
        sum[rt<<1|1] = (r-m)*cov[rt];
        cov[rt<<1] = cov[rt<<1|1] = cov[rt];
        cov[rt] = -1;
    }
}
void pushUP(int rt){
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void update(int L,int R,int c,int l,int r,int rt){
    if(L<=l&&R>=r){
        cov[rt] = c;
        sum[rt] = (r-l+1)*c;
        return;
    }
    pushDOWN(rt,l,r);
    int m = (l+r)>>1;
    if(m>=L)update(L,R,c,lson);
    if(m<R)update(L,R,c,rson);
    pushUP(rt);
}
int query(int L,int R,int l,int r,int rt){
    if(L<=l&&R>=r){
        return sum[rt];
    }
    pushDOWN(rt,l,r);
    int m = (l+r)>>1,ret = 0;
    if(m>=L)ret+=query(L,R,lson);
    if(m<R)ret+=query(L,R,rson);
    pushUP(rt);
    return ret;
}
int gcd(int a,int b){
    return ((a%b)==0)?b:gcd(b,a%b);
}
void print(int cas,int mo,int so){
    int div = gcd(mo,so);
    mo/=div;so/=div;
    if(so==1){
        printf("%d\n",mo);
    }else if(mo!=0){
        printf("%d/%d\n",mo,so);
    }else{
        printf("%d\n",0);
    }
}
int main(){
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++){
        scanf("%d%d",&n,&q);
        build();
        printf("Case %d:\n",cas);
        while(q--){
            int a,b,c,d;
            scanf("%d",&a);
            if(a==1){
                scanf("%d%d%d",&b,&c,&d);
                update(b,c,d,0,n-1,1);
            }else{
                scanf("%d%d",&b,&c);
                int mo = query(b,c,0,n-1,1);
                print(cas,mo,(c-b+1));
            }
        }
    }
    return 0;
}