2017"百度之星"程序设计大赛 - 初赛（B）_百度之星程序设计大赛初赛csdn-优快云博客

本文介绍了两种算法问题：一是通过动态规划求解特定条件下的组合数；二是解决区间覆盖问题，采用排序与滑动窗口的方法找到最优解。

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1001

推了推之后发现就是求组合数了

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e3 + 10;
const int INF = 1e9 + 10;
LL dp[qq][qq];
void Init() {
	for(int i = 1; i <= 1000; ++i) {
		for(int j = 1; j <= i; ++j) {
			if(i == 1 || j == 1)	dp[i][j] = max(i, j);
			else	dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j]) % MOD;
		}
	}
}

int main(){
	Init();
	int t;	scanf("%d", &t);
	while(t--) {
		int a, b;	scanf("%d%d", &a, &b);
		if(a >= b)	printf("%lld\n", dp[a][b]);
		else	printf("%lld\n", dp[b][a]);	
	}
	return 0;
}

1006

这题模拟一下、排序，首先把所有覆盖的区间找出来，然后利用尺取法找出最大值

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 2e5 + 10;
const int INF = 1e9 + 10;
struct Node {
	int l, r;
	bool operator < (const Node &a) const {
		if(l == a.l)	return r > a.r;
		return l < a.l;
	}
}x[qq], y[qq];
int n, m;

int main(){
	while(scanf("%d%d", &n, &m) != EOF) {
		for(int i = 0; i < n; ++i) {
			scanf("%d%d", &x[i].l, &x[i].r);
		}
		sort(x, x + n);
		int l = x[0].l, r = x[0].r;
		int cnt = 0;
		for(int i = 1; i < n; ++i) {
			if(x[i].l <= r) {
				r = max(r, x[i].r);
			} else {
				y[cnt].l = l, y[cnt].r = r;
				l = x[i].l, r = x[i].r;
				cnt++;
			}
		}
		y[cnt].l = l, y[cnt].r = r;
		cnt++;
		l = y[0].l, r = y[0].r;
		int p = 0;
		int maxn = r - l + 1 + m;
		for(int i = 1; i < cnt; ++i) {
			int dis = y[i].l - y[i - 1].r - 1;
			if(m >= dis) {
				m -= dis;
				r = y[i].r;
				maxn = max(maxn, r - l + 1 + m);
			} else {
				while(p < i && m < dis) {
					m += y[p + 1].l - y[p].r - 1;
					l = y[p + 1].l;
					p++;
				}
				if(m >= dis)	m -= dis;
				r = y[i].r;
//a				printf("%d %d\n", l, r);
				maxn = max(maxn, r - l + 1 + m);
			}
		}
		printf("%d\n", maxn);
	}
	return 0;
}