BZOJ1583 USACO 2009 Mar Gold 1.Moon Mooing Solution

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题意：自行脑补。

Sol:一种显然的方法就是拿堆维护前n小元素，时间复杂度O(nlogn).不过这样会超时。

事实上由于两个函数都是单调递增的，我们用两个指针在同一个序列中进行线性合并并添加元素，不难证明时间复杂度为O(n).

Code:

Heap O(nlogn)

#include <cstdio>#include <cstring>#include <cctype>#include <iostream>#include <algorithm>#include <queue>#include <vector>#include <functional>using namespace std;typedef unsigned long long LU;struct Case {LU x;Case(LU _x = 0):x(_x){}bool operator > (const Case &B) const {return x < B.x;}bool operator < (const Case &B) const {return x > B.x;}};priority_queue<Case> q;int main() {int c, N;scanf("%d%d", &c, &N);register int i, j;int a1, b1, d1;scanf("%d%d%d", &a1, &b1, &d1);int a2, b2, d2;scanf("%d%d%d", &a2, &b2, &d2);q.push(Case(c));LU x, last = 0;for(i = 1; i <= N;) {x = q.top().x;q.pop();if (x != last) {q.push(Case(x * a1 / (LU)d1 + (LU)b1));q.push(Case(x * a2 / (LU)d2 + (LU)b2));last = x;if (i == N)printf("%llu", x);++i;}}return 0;}

Merge Sort O(n)

#include <cstdio>#include <cstring>#include <climits>#include <algorithm>#include <iostream>using namespace std;typedef unsigned long long LU;LU seq[4000010];int main() {int c, N;scanf("%d%d", &c, &N);register int i, j;int a1, b1, d1;scanf("%d%d%d", &a1, &b1, &d1);int a2, b2, d2;scanf("%d%d%d", &a2, &b2, &d2);seq[1] = c;int p1 = 1, p2 = 1;LU nxt1 = seq[1] * a1 / d1 + b1;LU nxt2 = seq[1] * a2 / d2 + b2;for(i = 2; i <= N; ) {if (nxt1 < nxt2) {if (nxt1 != seq[i - 1])seq[i++] = nxt1;nxt1 = seq[++p1] * a1 / d1 + b1;}else {if (nxt2 != seq[i - 1])seq[i++] = nxt2;nxt2 = seq[++p2] * a2 / d2 + b2;}}printf("%llu", seq[N]);return 0;}

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