離散化+排序

離散化統計人數就好，本來不難，但是測試點太喪心病狂了。。。CF還是大哥啊

#include <bits/stdc++.h> #define INF 0x3f3f3f3f using namespace std; typedef long long ll; inline int lowbit(int x){ return x & (-x); } inline int read(){int X = 0, w = 0; char ch = 0;while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();return w ? -X : X; } inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; } inline int lcm(int a, int b){ return a / gcd(a, b) * b; } template<typename T> inline T max(T x, T y, T z){ return max(max(x, y), z); } template<typename T> inline T min(T x, T y, T z){ return min(min(x, y), z); } template<typename A, typename B, typename C> inline A fpow(A x, B p, C yql){A ans = 1;for(; p; p >>= 1, x = 1LL * x * x % yql)if(p & 1)ans = 1LL * x * ans % yql;return ans; } const int N = 200005; int a[N], b[N<<2], tot, num[N<<2]; struct Film{int x, y, cnt1, cnt2, id;bool operator < (const Film &rhs) const {if(cnt1 != rhs.cnt1) return cnt1 > rhs.cnt1;if(cnt2 != rhs.cnt2) return cnt2 > rhs.cnt2;return id < rhs.id;} }film[N]; int main(){int n = read();for(int i = 0; i < n; i ++) a[i] = read(), b[++tot] = a[i];int m = read();for(int i = 0; i < m; i ++) film[i].x = read(), b[++tot] = film[i].x, film[i].id = i;for(int i = 0; i < m; i ++) film[i].y = read(), b[++tot] = film[i].y;sort(b + 1, b + tot + 1);tot = unique(b + 1, b + tot + 1) - b - 1;for(int i = 0; i < n; i ++){num[lower_bound(b + 1, b + tot + 1, a[i]) - b] ++;}for(int i = 0; i < m; i ++){film[i].cnt1 = num[lower_bound(b + 1, b + tot + 1, film[i].x) - b];film[i].cnt2 = num[lower_bound(b + 1, b + tot + 1, film[i].y) - b];}sort(film, film + m);printf("%d\n", film[0].id + 1);return 0; }

轉載于:https://www.cnblogs.com/onionQAQ/p/10533850.html

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