差分例题-增减序列

Posted 2021-05-15 你可以当我没名字

https://www.acwing.com/problem/content/102/

在[l,r]区间整体增加或减少可以想到使用差分,

依照题意即使差分数组全部为0

所以我们可以在差分数组里两两相对,正数–,负数++

最后剩下的可能不是0,所以最后那个不等于0的数可以切换到cb[1]或者cb[n+1] (即开始和最后,因为最开始cb[0]和cb[n+1]是0,差分之后可以算到这里)

为了方便处理数据下标从一开始

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<iostream>
#include<sstream>
#include<set>
#include<map>
#include<queue>
#include<bitset>
#include<vector>
#include<limits.h>
#include<assert.h>
#include<functional>
#include<numeric>
#include<ctime>
#define ini(a) memset(a,0,sizeof(a))
#define ini2(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define lowbit(x) x&(-x)
#define DBG(x) \\
  (void)(cout << "L" << __LINE__ \\
  << ": " << #x << " = " << (x) << '\\n')
#define TIE \\
	cin.tie(0);cout.tie(0)\\
	ios::sync_with_stdio(false);
#define sc ll T;cin>>T;while(T--)
#define IN freopen("in.txt","r",stdin);
#define OUT srand((unsigned)time(NULL)); \\
	freopen("out.txt","w",stdout);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int maxn = 500000;
ll cb[maxn];

ll fastpower(ll a,ll b,ll p){
	ll res = 1;
	while(b)
    {
        if(b&1) res=res*a%p;
        b>>=1;
        a=a*a%p;
    }
    return res%p;
}


int main()
{
	ini(cb);
	ll n;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>cb[i];
	}
	
	for(int i=n;i>=1;i--){//cb[0]是0,所以i>=1或者i>=2都行
		cb[i] -= cb[i-1];
	}
	
	ll a = 0,b = 0;
	for(int i=2;i<=n;i++){//这里就要除去差分数组cb[1],理由在上面
		if(cb[i]>0)a+=cb[i];
		else b-=cb[i];
	}
	
	cout<<min(a,b)+abs(a-b)<<"\\n";
	cout<<abs(a-b)+1; 
	
	return 0;
}