牛客巅峰赛第11场代码
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define debug(x) cerr<<#x<<" = "<<x
#define sp <<" "
#define el <<endl
#define fgx cerr<<" ---------------------- "<<endl
#define uint unsigned int
#define ULL unsigned long long
#define DB double
#define LDB long double
#define pii pair<int,int>
#define mpt make_pair
#define pb push_back
#define fr first
#define sc second
#define M 1000020//Size
#define INF 1000000000
#define INFLL 1000000000000000000
inline int read(){
int nm=0,fh=1; char c=getchar();
for(;!isdigit(c);c=getchar()) if(c=='-') fh=-1;
for(;isdigit(c);c=getchar()) nm=nm*10+c-'0';
return nm*fh;
}
#define mod 1000000007//About
inline int add(int x,int y){return (x+y>=mod)?(x+y-mod):(x+y);}
inline int mns(int x,int y){return (x-y<0)?(x-y+mod):(x-y);}
inline int mul(LL x,LL y){return x*y%mod;}
inline void upd(int &x,int y){x=(x+y>=mod)?(x+y-mod):(x+y);}
inline void dec(int &x,int y){x=(x-y<0)?(x-y+mod):(x-y);}
inline int qpow(int x,LL sq){int res=1;for(;sq;sq>>=1,x=mul(x,x))if(sq&1)res=mul(res,x);return res;}
int n,ans[M],tp;
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
* @param n int整型 节点个数
* @param u int整型vector
* @param v int整型vector
* @return int整型
*/
int oddnumber(int n, int m, vector<int>& l, vector<int>& r) {
// write code here
for(int i=1;i<=m;i++) ans[l[i-1]]++,--ans[r[i-1]+1];
for(int i=1;i<=n;i++) ans[i]+=ans[i-1];
for(int i=1;i<=n;i++) ans[i]+=m;
int res=0;
for(int i=1;i<=n;i++) if(ans[i]&1) res++;
return res;
}
}t;
#牛客编程巅峰赛##题解#
