SPOJ CRAN02 - Roommate Agreement

题目链接http://www.spoj.com/problems/CRAN02/

题目大意:N个数字组成的序列,和为0的连续子序列的个数。N<1e6

解题思路:计算前缀和,统计每个数字出现的次数,那么对于数字sum[i], 如果存在k个sum[i],则代表有C(k, 2)个序列和为0,而如果sum[i] = 0,则还要累加上对应的k值。

代码:

 1 ll n;
 2 int a[maxn];
 3 ll sum[maxn];
 4 map<int, int> mmp;
 5
 6 void solve(){
 7     memset(sum, 0, sizeof(sum));
 8     for(int i = 1; i <= n; i++) {
 9         sum[i] = sum[i - 1] + a[i];
10         mmp[sum[i]]++;
11     }
12     ll ans = 0;
13     for(int i = 1; i <= n; i++){
14         ll k = mmp[sum[i]];
15         if(sum[i] == 0) ans += k;
16         mmp[sum[i]] = 0;
17         if(k >= 2) ans += k * (k - 1) / 2;
18     }
19     printf("%lld\n",ans);
20 }
21 int main(){
22     int t;
23     scanf("%d", &t);
24     for(int i = 1; i <= t; i++){
25         mmp.clear();
26         scanf("%lld", &n);
27         for(int j = 1; j <= n; j++)
28             scanf("%d", &a[j]);
29         solve();
30     }
31 }

题目:

CRAN02 - Roommate Agreement

#math #ad-hoc-1

Leonard was always sickened by how Sheldon considered himself better than him. To decide once and for all who is better among them they decided to ask each other a puzzle. Sheldon pointed out that according to Roommate Agreement Sheldon will ask first. Leonard seeing an opportunity decided that the winner will get to rewrite the Roommate Agreement.

Sheldon thought for a moment then agreed to the terms thinking that Leonard will never be able to answer right. For Leonard, Sheldon thought of a puzzle which is as follows. He gave Leonard n numbers, which can be both positive and negative. Leonard had to find the number of continuous sequence of numbers such that their sum is zero.

For example if the sequence is- 5, 2, -2, 5, -5, 9

There are 3 such sequences

2, -2

5, -5

2, -2, 5, -5

Since this is a golden opportunity for Leonard to rewrite the Roommate Agreement and get rid of Sheldon‘s ridiculous clauses, he can‘t afford to lose. So he turns to you for help. Don‘t let him down.

Input

First line contains T - number of test cases

Second line contains n - the number of elements in a particular test case.

Next line contain n elements, ai  (1<=i<= n) separated by spaces.

Output

The number of such sequences whose sum if zero.

Constraints

1<=t<=5

1<=n<=10^6

-10<= ai <= 10

Example

Input:

2

4

0 1 -1 0

6

5 2 -2 5 -5 9

Output:

63
时间: 2024-12-29 11:55:06

SPOJ CRAN02 - Roommate Agreement的相关文章

SPOJ-CRAN02 - Roommate Agreement(前缀和)

CRAN02 - Roommate Agreement Leonard was always sickened by how Sheldon considered himself better than him. To decide once and for all who is better among them they decided to ask each other a puzzle. Sheldon pointed out that according to Roommate Agr

SPOJ 705 Distinct Substrings(后缀数组)

[题目链接] http://www.spoj.com/problems/SUBST1/ [题目大意] 给出一个串,求出不相同的子串的个数. [题解] 对原串做一遍后缀数组,按照后缀的名次进行遍历, 每个后缀对答案的贡献为n-sa[i]+1-h[i], 因为排名相邻的后缀一定是公共前缀最长的, 那么就可以有效地通过LCP去除重复计算的子串. [代码] #include <cstdio> #include <cstring> #include <algorithm> usi

SPOJ 3273

传送门: 这是一道treap的模板题,不要问我为什么一直在写模板题 依旧只放代码 1 //SPOJ 3273 2 //by Cydiater 3 //2016.8.31 4 #include <iostream> 5 #include <cstring> 6 #include <ctime> 7 #include <cmath> 8 #include <cstdlib> 9 #include <string> 10 #include

spoj GCJ1C09C Bribe the Prisoners

题目链接: http://www.spoj.com/problems/GCJ1C09C/ 题意: In a kingdom there are prison cells (numbered 1 to P) built to form a straight line segment. Cells number i and i+1 are adjacent, and prisoners in adjacent cells are called "neighbours." A wall wi

SPOJ QTREE Query on a tree ——树链剖分 线段树

[题目分析] 垃圾vjudge又挂了. 树链剖分裸题. 垃圾spoj,交了好几次,基本没改动却过了. [代码](自带常数,是别人的2倍左右) #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define maxn 20005 int T,n,fr[maxn],h[maxn],to[maxn],ne[maxn]

BZOJ 2588: Spoj 10628. Count on a tree 主席树+lca

2588: Spoj 10628. Count on a tree Description 给定一棵N个节点的树,每个点有一个权值,对于M个询问(u,v,k),你需要回答u xor lastans和v这两个节点间第K小的点权.其中lastans是上一个询问的答案,初始为0,即第一个询问的u是明文. Input 第一行两个整数N,M. 第二行有N个整数,其中第i个整数表示点i的权值. 后面N-1行每行两个整数(x,y),表示点x到点y有一条边. 最后M行每行两个整数(u,v,k),表示一组询问.

BZOJ 1002 + SPOJ 104 基尔霍夫矩阵 + 一个递推式。

BZOJ 1002 高精度 + 递推 f[1] = 1; f[2] = 5; f[i] = f[i - 1] * 3 - f[i - 2] + 2; SPOJ 104 裸 + 不用Mod 1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <algorithm> 5 #include <iostream> 6 7 using namespace std;

SPOJ QTREE 系列解题报告

题目一 : SPOJ 375 Query On a Tree http://www.spoj.com/problems/QTREE/ 给一个树,求a,b路径上最大边权,或者修改a,b边权为t. 1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <cstdlib> 5 #include <algorithm> 6 7 using namespace s

BZOJ 2226: [Spoj 5971] LCMSum( 数论 )

∑lcm(i,n) = ∑ i*n/(i,n) = ∑d|n∑(x,n)=d x*n/d = ∑d|n∑(t,n/d)=1t*n = n∑d|nf(d). f(d)表示1~d中与d互质的数的和, 即f(d) = d*φ(d)/2(d>=2). 然后O(n)筛φ, 每次询问暴力算即可...最大是100w,sqrt(100w)=1000内的质数是168个, 所以复杂度是O(n + T*168), 可以AC  ----------------------------------------------