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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#137435#2356. Partition of QueriesTeam_name#WA 2ms7764kbC++201.9kb2023-08-10 12:39:322023-08-10 12:39:35

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-08-10 12:39:35]
  • 评测
  • 测评结果:WA
  • 用时:2ms
  • 内存:7764kb
  • [2023-08-10 12:39:32]
  • 提交

answer

#pragma GCC optimize("Ofast")

#include <bits/stdc++.h>

#define _debugVar(x) { cerr << "* "#x" = " << x << endl; }
#define _debugHere { fprintf(stderr, "* Passing [%s] in Line %d\n", __FUNCTION__, __LINE__); }

using namespace std;
using LL = long long;
using PII = pair<int, int>;

const int N = 1e6+5;

int n;
LL m;
char str[N];
LL cnta[N], cntq[N], sum[N];
LL f[N];

double slope(int i, int j)
{
	static const auto X = [&](int j) -> double {
		return (double)cnta[j];
	};
	static const auto Y = [&](int j) -> double {
		return (double)cnta[j]*cntq[j]-sum[j]+f[j];
	};

	// fprintf(stderr, "* X(%d) = %.6lf\n", i, X(i));
	// fprintf(stderr, "* Y(%d) = %.6lf\n", i, Y(i));
	// fprintf(stderr, "* X(%d) = %.6lf\n", j, X(j));
	// fprintf(stderr, "* Y(%d) = %.6lf\n", j, Y(j));
	double res = (Y(j)-Y(i))/(X(j)-X(i));
	// fprintf(stderr, "slope(%d, %d) = %.6lf\n", i, j, res);
	return res;
}

int main()
{
	// freopen("1.in", "r", stdin);
	scanf("%d%lld", &n, &m);
	scanf(" %s", str+1);

	for(int i = 1; i <= n; i++) {
		cnta[i] = cnta[i-1]+(str[i] == '+');
		cntq[i] = cntq[i-1]+(str[i] == '?');
		sum[i] = sum[i-1]+(str[i] == '?')*cnta[i];
	}

	static int q[N];
	int hh = 1, tt = 0;
	f[0] = -m;
	q[++tt] = 0;

	for(int i = 1; i <= n; i++) {
		//_debugVar(slope(q[hh], q[hh+1]));
		// int j = hh;
		// while(j < tt && slope(q[j], q[j+1]) >= cntq[i]) j++;
		// while(hh < tt && slope(q[tt-1], q[tt]) <= cntq[i]) tt--;
		while(hh < tt && slope(q[hh], q[hh+1]) <= cntq[i]) hh++;
		int j = q[hh];
		f[i] = f[j]+(sum[i]-sum[j])-(cntq[i]-cntq[j])*cnta[j]+m;
		// fprintf(stderr, "f[%d] = %lld from %d\n", i, f[i], j);
		while(hh < tt && slope(q[tt-1], q[tt]) >= slope(q[tt-1], i)) tt--;
		q[++tt] = i;
	}

	// for(int i = 1; i <= n; i++) {
	// 	for(int j = 0; j <= i-1; j++) {
	// 		f[i] = min(f[i], f[j]+(sum[i]-sum[j])-(cntq[i]-cntq[j])*cnta[j]+m);
	// 	}
	// }
	printf("%lld\n", f[n]);
	return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 2ms
memory: 7764kb

input:

6 5
++??+?

output:

6

result:

ok single line: '6'

Test #2:

score: 0
Accepted
time: 1ms
memory: 5896kb

input:

6 8
++??+?

output:

7

result:

ok single line: '7'

Test #3:

score: 0
Accepted
time: 2ms
memory: 7700kb

input:

5 1
+++++

output:

0

result:

ok single line: '0'

Test #4:

score: -100
Wrong Answer
time: 1ms
memory: 7744kb

input:

10 0
++?+??++??

output:

8

result:

wrong answer 1st lines differ - expected: '0', found: '8'