Codeforces Round #353 (Div. 2) C 思维

Problem:

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给你一个数列,本身是一个环且和为0,每次可以执行一个操作就是相邻的成员之间传递值,
问让该环所有成员的值都变为0,需要执行最少多少次操作?

Analyse:

• 可以转化为求环中最多和为0的块的个数.
• 然后因为没一个块的和都为0,所以在两个相邻的和为0的块之间该数列的前缀和是相同的,我们要求最多的块数,
  只需要求最大的相同前缀和的频率即可
  Get:
  比赛的时候已经推出了第一步,不错,应该多思考,多发些性质,说不定就找到关键性质了.

/**********************jibancanyang************************** *Author*        :jibancanyang *Created Time*  : 二  5/17 23:44:56 2016**Code**:***********************[e-mail]**********************/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;vector<int> vi;#define pr(x) cout << #x << ": " << x << "  " #define pl(x) cout << #x << ": " << x << endl;#define pri(a) printf("%d\n",(a));#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%lld", &(n))#define sai(n) scanf("%I64d", &(n))#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++) const int mod = int(1e9) + 7, INF = 0x3f3f3f3f;const int maxn = 1e5 + 13;int main(void){#ifdef LOCAL    //freopen("/Users/zhaoyang/in.txt", "r", stdin);    //freopen("/Users/zhaoyang/out.txt", "w", stdout);#endif    map<ll, ll> mp;    int n;    cin >> n;   ll mins = 0, sum = 0, x;    for (int i = 0; i < n; i++) {        cin >> x;        mp[sum += x]++;        mins = max(mins, mp[sum]);    }    cout << n - mins << endl;    return 0;}