MemSQL Start[c]UP 2.0 - Round 2 - Online Round

题目链接

不得不说这场真心不好打啊...

A：黄金分割进制数，满足一个性质，对于第i位xi=xi?1+xi?2，这样一来就可以把所有位推到前两位去比较大小，不过单单这样直接搞果断爆longlong无限WA8，最后发现在推的过程中，有一位上面差值大于1，就可以直接判断了

B：不得不吐槽一下毛子的英语，Today‘s ying yu is very hao，题目看了非常非常久，不能更逗。其实看懂的就挺简单了，只要贪心分别考虑A放入B和B放入A的情况，然后判断是用复制好还是移动好即可

C：三分，设f(x)为其他人选票都小于x，自己选票大于等于x需要的开销，这样一开很容易证明f(x)是一个下凹的单峰函数，就可以用三分法求解了

代码：

A:

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const double eps = 1e-10;
const int N = 100005;

char A[N], B[N];
long long a[N], b[N];

int main() {
    scanf("%s%s", A, B);
    int an = strlen(A);
    int bn = strlen(B);
    for (int i = 0; i <  an; i++)
	a[i] = A[an - i - 1] - '0';
    for (int i = 0; i < bn ; i++)
	b[i] = B[bn - i - 1] - '0';
    int n = max(an, bn);
    for (int i = n - 1; i >= 2; i--) {
	if (a[i] >= b[i] ){
	    a[i] -= b[i];
	    b[i] = 0;
	}
	else if (a[i] <= b[i]) {
	    b[i] -= a[i];
	    a[i] = 0;
	}
	if (a[i] >= 2) {
	    printf(">\n");
	    return 0;
	}
	if (b[i] >= 2) {
	    printf("<\n");
	    return 0;
	}
	a[i - 1] += a[i]; a[i - 2] += a[i];
	b[i - 1] += b[i]; b[i - 2] += b[i];
    }
    long long aa = a[0], bb = a[1];
    long long cc = b[0], dd = b[1];
    long long x = 2 * (aa - cc) - (dd - bb);
    long long y = (dd - bb);
    if (x < 0 && y > 0) {
	printf("<\n");
    }
    else if (x > 0 && y < 0) {
	printf(">\n");
    }
    else if (x >= 0 && y >= 0) {
	x = x * x;
	y = y * y * 5;
	if (x == y) printf("=\n");
	else if (x < y) printf("<\n");
	else printf(">\n");
    }
    else {
	x = x * x;
	y = y * y * 5;
	if (x == y) printf("=\n");
	else if (x < y) printf(">\n");
	else printf("<\n");
    }
    return 0;
}

B:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef unsigned long long ll;

const int N = 100005;

ll a[N], b[N], suma, sumb, n, m;

int main() {
    scanf("%llu%llu", &n, &m);
    for (int i = 0; i < n; i++) {
	scanf("%llu", &a[i]);
	suma += a[i];
    }
    for (int i = 0; i < m; i++) {
	scanf("%llu", &b[i]);
	sumb += b[i];
    }
    sort(a, a + n);
    sort(b, b + m);
    ll ans = 10000000000000000000ULL;
    ll sum = 0;
    for (ll i = 0; i < n - 1; i++) {
	if (sumb <= a[i]) {
	    sum += (n - i) * sumb;
	    ans = min(ans, sum);
	    break;
	}
	sum += a[i];
    }
    ans = min(ans, sum + sumb);
    sum = 0;
    for (ll i = 0; i < m - 1; i++) {
	if (suma <= b[i]) {
	    sum += (m - i) * suma;
	    ans = min(ans, sum);
	}
	sum += b[i];
    }
    ans = min(ans, sum + suma);
    printf("%llu\n", ans);
    return 0;
}

C:

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 100005;

int n, save[N];
vector<int> g[N];

int cal(int x) {
    int ans = 0, have = g[0].size(), sn = 0;
    for (int i = 1; i <= 100000; i++) {
	int j = 0;
	if (x <= g[i].size()) {
	    for (; j < g[i].size() - x + 1; j++) {
		ans += g[i][j];
		have++;
	    }
	}
	for (; j < g[i].size(); j++)
	    save[sn++] = g[i][j];
    }
    sort(save, save + sn);
    for (int i = 0; i < x - have; i++)
	ans += save[i];
    return ans;
}

int main() {
    scanf("%d", &n);
    int a, b;
    for (int i = 0; i < n; i++) {
	scanf("%d%d", &a, &b);
	g[a].push_back(b);
    }
    for (int i = 1; i <= 100000; i++)
	sort(g[i].begin(), g[i].end());
    int l = 1, r = n;
    while (l < r - 2) {
	int midl = (l * 2 + r) / 3;
	int midr = (l + r * 2) / 3;
	if (cal(midl) > cal(midr)) l = midl;
	else r = midr;
    }
    int ans = INF;
    for (int i = l; i <= r; i++)
	ans = min(ans, cal(i));
    printf("%d\n", ans);
    return 0;
}

MemSQL Start[c]UP 2.0 - Round 2 - Online Round A,B,C