Educational Codeforces Round 41 E. Tufurama (961E)

【题解】

　　第一眼看题飞快地想到一种做法，然后假掉了。

　　这道题其实是主席树的模板题来着。但是也有别的水法。

　　我们可以发现每个位置的查询区间是[1,min(a[i],i-1)]，所以我们可以把查询区间按照右端点排序。开一个权值树状数组记录前i个a[i]的出现情况。我们从1到n按顺序插入a[i]，每个位置上都统计min(a[j],j-1)=i的j的答案。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #define rg register
 4 #define N 200010
 5 using namespace std;
 6 int n,a[N],b[N],d[N],t[N];
 7 long long ans;
 8 inline int read(){
 9     int k=0,f=1; char c=getchar();
10     while(c<'0'||c>'9')c=='-'&&(f=-1),c=getchar();
11     while('0'<=c&&c<='9')k=k*10+c-'0',c=getchar();
12     return k*f;
13 }
14 inline void add(int x){
15     for(;x<=n;x+=(x&-x)) t[x]++;
16 }
17 inline int query(int x){
18     int ret=0; for(;x>0;x-=(x&-x)) ret+=t[x]; return ret;
19 }
20 inline bool cmp(int x,int y){
21     return d[x]<d[y];
22 }
23 int main(){
24     n=read();
25     for(rg int i=1;i<=n;i++) a[i]=min(read(),n),b[i]=i,d[i]=min(a[i],i-1);
26     sort(b+1,b+1+n,cmp);
27     for(rg int i=1,j=1;i<=n;i++){
28         add(a[i]);
29         while(d[b[j]]<i&&j<=n) j++;
30         while(j<=n&&d[b[j]]==i) ans+=i-query(b[j++]-1);
31     }
32     printf("%I64d\n",ans);
33     return 0;
34 }

View Code

转载于:https://www.cnblogs.com/DriverLao/p/8728504.html

Educational Codeforces Round 41 E. Tufurama (961E)相关推荐

Educational Codeforces Round 41 (Rated for Div. 2) ABCDEF
最近打的比较少...就只有这么点题解了. A. Tetris time limit per test 1 second memory limit per test 256 megabytes inpu ...
Educational Codeforces Round 41(已补D,E)
D. Pair Of Lines 题目大意: 给\(n\)个点,问是否存在两条直线(可以为同一条)能覆盖所有的点. 解题思路: 首先一个或者两个或者三个点都是肯定可以用两条线覆盖的. 如果点的个数大于 ...
Educational Codeforces Round 41 (Rated for Div. 2)
这个D可以参考这个B的思路 A. Tetris time limit per test 1 second memory limit per test 256 megabytes input stand ...
Educational Codeforces Round 41 (Rated for Div. 2) F. k-substrings
F. k-substrings time limit per test 4 seconds memory limit per test 256 megabytes input standard inp ...
Educational Codeforces Round 90 (Rated for Div. 2)（A, B, C, D, E）
Educational Codeforces Round 90 (Rated for Div. 2) Donut Shops 思路分三种情况: a==c/ba == c / ba==c/b这个时候两 ...
Educational Codeforces Round 24 E. Card Game Again(双指针)
题目链接:Educational Codeforces Round 24 E. Card Game Again 题意: 给你n个数和一个数k. 现在每次可以拿掉前x个数,后y个数,剩下的数的乘积要能被 ...
Educational Codeforces Round 138 (Rated for Div. 2) D
Educational Codeforces Round 138 (Rated for Div. 2) D. Counting Arrays 题意给定长度为nnn的数组aaa,若gcd(ai,i)= ...
Educational Codeforces Round 113 (Rated for Div. 2) ABCD 解题思路
Educational Codeforces Round 113 (Rated for Div. 2) 经典卡CCC秒DDD,可惜了怪自己特判写错了吧,对式子找了半天问题结果根本不是式子的问题 A ...
Educational Codeforces Round 112 (Rated for Div. 2)-A. PizzaForces-题解
目录 Educational Codeforces Round 112 (Rated for Div. 2)-A. PizzaForces Problem Description Input Outp ...

Educational Codeforces Round 41 E. Tufurama (961E)

Educational Codeforces Round 41 E. Tufurama (961E)相关推荐

最新文章

热门文章