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

1.      1 i j v - change the value of the elements from i^th index to j^th index to v.

2.      2 i j - find the average value of the integers from i^th index to j^th index.

You can assume that initially all the values in the array are 0.

Input

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

Each case contains two integers: n (1 ≤ n ≤ 10⁵), q (1 ≤ q ≤ 50000), where n denotes the size of the array. Each of the next q lines 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 from i to j. If the result is an integer, print it. Otherwise print the result in 'x/y' form, where x denotes the numerator and y denotes the denominator of the result and x and y are 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|1void 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;}