Codeforces 675C Money Transfers【贪心】
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题目链接:
http://codeforces.com/contest/675/problem/C
题意:
给定几个数,有正有负,每个数可以向相邻的数转移,问最少的转移次数使得最后所有数均为0。
分析:
我们可以将数列化为几个连续的区间,其中每个区间的数和为0,且在区间长度为K的区间中,操作数为K-1,我们就是要最大化这样的区间个数。可以维护一个前缀和,这样两个相同的前缀之间的区间和即为0。
代码:
/*On a hill is a tree, on a tree is a bough;My heart for the Lord, but he never knows.--Created by jiangyuzhu--2016/5/17*/#include<cstdio>#include<iostream>#include<queue>#include<cstring>#include<stack>#include<vector>#include<algorithm>#include<map>#include<set>#include<cmath>using namespace std;#define pr(x) cout << #x << ": " << x << " "#define pl(x) cout << #x << ": " << x << endl;#define sa(x) scanf("%d",&(x))#define sal(x) scanf("%I64d",&(x))#define mdzz cout<<"mdzz"<<endl;typedef long long ll;const int maxn = 1e5 + 5, mod = 1e9 + 7;ll a[maxn];map<ll, int>m;int main(void){ int n;sa(n); int ans = 0; ll tot = 0; for(int i = 0; i < n; i++){ sal(a[i]); tot += a[i]; m[tot]++; ans = max(ans, m[tot]); } printf("%d", n - ans); return 0;}
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