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 fromithindex tojthindex tov.
2.2 i j- find the average value of the integers fromithindex tojthindex.
You can assume that initially all the values in the array are0.
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)
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 |
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; }