题目链接
ac代码
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define mp make_pair #define pi acos(-1) #define pii pair #define pll pair #define ll long long #define ld long double #define MEMS(x) memset(x, -1, sizeof(x)) #define MEM(x) memset(x, 0, sizeof(x)) const int inf = 0x3f3f3f3f; const int maxn = 500001; using namespace std; int n; struct Intg { int data; int pos; bool operator<(const Intg num) const { return data < num.data; } }a[maxn]; int p[maxn], d[maxn]; int lowbit(int x) { return x & (-x); } void update(int x) { while(x <= maxn) { d[x]++; x += lowbit(x); } } int getsum(int x) { int ans = 0; while(x) //一定写成这种形式，否则写成for循环形式会导致超时 { ans += d[x]; x -= lowbit(x); } return ans; } int main() { // freopen("input.txt", "r", stdin); // freopen("output.txt", "w", stdout); while(scanf("%d", &n) != EOF && n) { for(int i = 1; i <= n; i++) { scanf("%d", &a[i].data); a[i].pos = i; } sort(a + 1, a + n + 1); p[a[1].pos] = 1; int idx = 1; for(int i = 2; i <= n ; i++) { if(a[i].data == a[i-1].data) p[a[i].pos] = idx; else p[a[i].pos] = ++idx; } ll ans = 0; memset(d, 0, sizeof(d)); for(int i = 1; i <= n; i++) { update(p[i]); ans += i - getsum(p[i]); } printf("%lld\n", ans); } }
