Codeforces Round #303 (Div. 2)

标签：acm   codeforces   

题目传送：Codeforces Round #303 (Div. 2)

这场题目比较简单，就最后一个题要好好想想

A. Toy Cars

思路：水

AC代码：

#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <cctype>
#define LL long long
#define INF 0x7fffffff
using namespace std;

int a[105];
int cnt;

int main() {
	int n;
	scanf("%d", &n);
	cnt = 0;
	for(int i = 0; i < n; i ++) {
		int flag = 1;
		for(int j = 0; j < n; j ++) {
			int t;
			scanf("%d", &t);
			if(t == 1 || t == 3) flag = 0;
		}
		if(flag) a[cnt ++] = i + 1;
	}
	printf("%d\n", cnt);
	for(int i = 0; i < cnt; i ++) {
		printf("%d ", a[i]);
	}
	return 0;
}

B. Equidistant String

题意：找一个字符串，使得他与给出的两个字符串的“距离相同”。具体做法看代码。

AC代码：

#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <cctype>
#define LL long long
#define INF 0x7fffffff
using namespace std;

char s1[100005], s2[100005];

int main() {
	scanf("%s %s", s1, s2);
	int len = strlen(s1);
	int cnt = 0;
	for(int i = 0; i < len; i ++) {
		if(s1[i] != s2[i]) cnt ++;
	}
	if(cnt & 1) {
		printf("impossible\n");
	}
	else {
		int m = cnt / 2;
		for(int i = 0; i < len; i ++) {
			if(s1[i] != s2[i]) {
				if(s1[i] == '1') s1[i] = '0';
				else s1[i] = '1';
				m --;
			}
			if(m == 0) break;
		}
		printf("%s\n", s1);
	}
	return 0;
}

C. Woodcutters

思路：砍树，尽量砍得最多，贪心，尽量往左边砍。

AC代码：

#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <cctype>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 100005;
int x[maxn], h[maxn];

int main() {
	int n;
	scanf("%d", &n);
	for(int i = 0; i < n; i ++) {
		scanf("%d %d", &x[i], &h[i]);
	}
	int ans = 1;
	int cur = -INF;
	for(int i = 0; i < n - 1; i ++) {
		if(i == 0) {
			ans ++;
			cur = x[i];
		}
		else if(x[i] - h[i] > cur && x[i] - h[i] > x[i - 1]) {
			ans ++;
			cur = x[i];
		}
		else if(x[i] > cur && x[i] + h[i] < x[i + 1]) {
			ans ++;
			cur = x[i] + h[i];
		}
	}
	printf("%d\n", ans);
	return 0;
}

D. Queue

思路：排下序然后贪心一下就ok了。

AC代码：

#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <cctype>
#define LL long long
#define INF 0x7fffffff
using namespace std;

int a[100005];

int main() {
	int n;
	scanf("%d", &n);
	for(int i = 0; i < n; i ++) {
		scanf("%d", &a[i]);
	}
	sort(a, a + n);
	LL ans = 1;
	LL sum = a[0];
	for(int i = 1; i < n; i ++) {
		if(a[i] >= sum) {
			ans ++;
			sum += a[i];
			//cout << sum << endl;
		}
	}
	cout << ans << endl;
	return 0;
}

E. Paths and Trees

待补...

Codeforces Round #303 (Div. 2)

标签：acm   codeforces   

原文地址：http://blog.csdn.net/u014355480/article/details/45877649