标签:open stdin mod lin 树状 c_str efi input ace
求逆序对,可以树状数组,但是这题n^2也能过。。。
1 #define mm(a) memset(a,0,sizeof(a)); 2 #define max(x,y) (x)>(y)?(x):(y) 3 #define min(x,y) (x)<(y)?(x):(y) 4 #define Fopen freopen("1.in","r",stdin); freopen("m.out","w",stdout); 5 #define rep(i,a,b) for(int i=(a);i<=(b);i++) 6 #define per(i,b,a) for(int i=(b);i>=(a);i--) 7 #include<bits/stdc++.h> 8 typedef long long ll; 9 #define PII pair<ll,ll> 10 using namespace std; 11 const int INF=0x3f3f3f3f; 12 const int MAXN=(int)2e5 + 5; 13 const ll mod=1e9+7; 14 15 16 string input,temp; 17 vector<int>v; 18 int n; 19 int main() { 20 while (cin >> input) { 21 istringstream iss(input); 22 v.push_back(0); 23 while (getline(iss, temp, ‘,‘)) { 24 v.push_back(atoi(temp.c_str())); 25 } 26 n=v.size()-1; 27 int cnt=0; 28 for(int i=1;i<=n;i++){ 29 for(int j=i+1;j<=n;j++){ 30 if(v[i]>v[j])cnt++; 31 } 32 } 33 printf("%d\n",cnt); 34 } 35 return 0; 36 }
标签:open stdin mod lin 树状 c_str efi input ace
原文地址:https://www.cnblogs.com/dogenya/p/10815597.html