今天是2019.5.27 距离NOIP2019还有165天
又是一道入门难度的题
这次没什么好说的 两分钟AC
O(n ^ 2)做法
想更快一点的话可以用树状数组求逆序对(比较懒没写)
代码
#include<cstdio> #include<cstring> #include<cmath> #include<cctype> #include<cstdlib> #include<string> #include<iostream> #include<algorithm> #include<set> #include<map> #include<queue> #include<stack> #include<vector> #define enter puts("") #define space putchar(' ') using namespace std; typedef long long ll; ll read() { ll op = 1, ans = 0; char ch = getchar(); while(ch < '0' || ch > '9') { if(ch == '-') op = 0; ch = getchar(); } while(ch >= '0' && ch <= '9') { ans *= 10; ans += ch - '0'; ch = getchar(); } return op ? ans : -ans; } void write(ll x) { if(x < 0) { x = -x; putchar('-'); } if(x >= 10) write(x / 10); putchar(x % 10 + '0'); } ll drug[105], zy[105], n; int main() { n = read(); for(int i = 1;i <= n;i++) { drug[i] = read(); for(int j = 1;j < i;j++) { if(drug[j] < drug[i]) zy[i]++; } } for(int i = 1;i <= n;i++) write(zy[i]), space; enter; return 0; }