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折半枚举

Posted by Ioencgc at 2023-05-02 11:53:05 on Problem 2785
#include<iostream>
#include<stack>
#include<string.h>
#include<cmath>
#include<iomanip>
#include<algorithm>
#include<climits>
#include<cstdio>
#include<vector>
#include<sstream>
#include<ctype.h>
#include<set>
#include<map>
#include<ctime>
#include<stdlib.h>
#include<queue>
#include<bitset>
#define usi unsigned int
#define ull unsigned long long
using namespace std;
template<typename TTT>inline void mr(TTT& theNumberToRead)
{
	theNumberToRead = 0;
	bool prn = false;
	char c = getchar();
	while (!isdigit(c))
	{
		if (c == '-')prn = true;
		c = getchar();
	}
	while (isdigit(c))
		theNumberToRead = 10 * theNumberToRead + (c ^ 48), c = getchar();
	if (prn)
		theNumberToRead = -theNumberToRead;
}
template<typename TTT>inline TTT mrr()
{
	TTT theNumberToRead = 0;
	bool prn = false;
	char c = getchar();
	while (!isdigit(c))
	{
		if (c == '-')prn = true;
		c = getchar();
	}
	while (isdigit(c))
		theNumberToRead = 10 * theNumberToRead + (c ^ 48), c = getchar();
	return prn ? -theNumberToRead : theNumberToRead;
}
template<typename T>void my_write(T x)
{
	if (x)
		my_write(x / 10),
		putchar(x % 10 ^ 48);
}
template<typename T>void mw(T x, char mid)
{
	if (x)
	{
		if (x < 0)
			putchar('-'),
			x = -x;
		my_write(x);
	}
	else putchar(48);
	if (mid)
		putchar(mid);
}
// ******************************************华丽的分割线******************************************
// ******************************************华丽的分割线******************************************
// ******************************************华丽的分割线******************************************
// ******************************************华丽的分割线******************************************
// ******************************************华丽的分割线******************************************
const int maxn = 4000;
int A[maxn], B[maxn], C[maxn], D[maxn], CD[maxn * maxn];
int main()
{
	int n;
	mr(n);
	for (int i = 0; i < n; ++i)
		mr(A[i]), mr(B[i]), mr(C[i]), mr(D[i]);
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < n; ++j)
			CD[i * n + j] = C[i] + D[j];
	sort(CD, CD + n * n);
	long long ans = 0;
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < n; ++j)
			ans += upper_bound(CD, CD + n * n, -A[i] - B[j]) - 
			lower_bound(CD, CD + n * n, -A[i] - B[j]);
	mw(ans, '\n');

	return 0;
}

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