Educational Codeforces Round 55 (Rated for Div. 2)

A. Vasya and Book

题意：

给四个数据 n, x, y, d。 n为书总共多少页，x为当前页数，y为目标页数，一次必须翻d页，但是范围在1~n，询问最少翻的次数

情况

1. abs(x-y)%d == 0 正好到达 则输出 abs(x-y)/d

2.往前翻到 1 此时为1 到 y 。 当 (y-1)%d == 0 则说明从1 可以翻到y 

3. 往后翻到n 此时为n 到y。 当（n-y) %d == 0 说明可以从n翻到y

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
#include<cstdlib>
#include<vector>
#include<queue>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;

int main() {
	int T, n, x, y, d;
	scanf("%d", &T);
	while(T--) {
		int ans = inf;
		scanf("%d%d%d%d", &n,&x,&y,&d);
		if(abs(x-y)%d == 0) { printf("%d
", abs(x-y)/d); continue ;}
		if((y-1)%d == 0) ans = min(ans, (x-1)/d+1 + (y-1)/d);
		if((n-y)%d == 0) ans = min(ans, + (n-x)/d + 1 + (n-y)/d);
		if(ans == inf) printf("-1
");
		else printf("%d
", ans);
	}
}

B. Vova and Trophies

题意：给n的字符的字符串，只有s与g，交换两个字母，让g的连续长度最长。输出最长的值

思路：最长肯定是GGGGSGGGGG这种最长，我们就记录一个S前后的g的长度。前为a后为b则答案为a+b+1

           特殊情况就是没有交换的S，我们的S必须交换成G。 如GGSGG 则答案为4不为5 a+b+1 = 2 + 2 + 1

代码

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
#include<cstdlib>
#include<vector>
#include<queue>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int MaxN = 1e5 + 5;
char s[MaxN];

int main() {
	int n;
	scanf("%d%s", &n, s);
	int a = 0, b = 0, c = 0, ans = 0;
	for(int i = 0; i < n; i++) {
		if(s[i] == 'G') a++, c++;
		else {
			b = a;
			a = 0;
		}
		ans = max(ans, a + b + 1);
	}
	ans = min(c, ans);  //ans最大也不能超过G的数量
	printf("%d
", ans);
}

C. Multi-Subject Competition

前缀和。。有点意思把

const int MaxN = 1e5 + 5;
int ans[MaxN];
vector<int> v[MaxN];

bool cmp(int a, int b) { return a > b; }

int main() {
	int n, m;
	scanf("%d%d", &n, &m);
	for(int i = 0; i < n; i++) {
		int s, r;
		scanf("%d%d", &s, &r);
		v[s].push_back(r);
	}

	int cnt = -1; // 记录v[i].size()的最大值 
	for(int i = 1; i <= m; i++) {
		int len = v[i].size(), sum = 0;
		sort(v[i].begin(), v[i].end(), cmp);
		cnt = max(cnt, len);

		for(int j = 0; j < len; j++) {
			sum += v[i][j];
			if(sum < 0) break;
			ans[j+1] += sum;
		}
	}
	sort(ans + 1, ans + 1 + cnt);
	printf("%d
", ans[cnt]);

}

D. Maximum Diameter Graph

建立一个直径最大的数。度大于1 的建立直径，然后将1添上

一个错误的答案。求一个有缘人

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<iostream>
#include<string>
#include<queue>
using namespace std;

int ans[510][510];
struct node{
	int val;
	int id;
}arr[510];

bool cmp(node a, node b) { return a.val < b.val; }

int main() {
	int n, num1 = 0, cnt = 0;
	scanf("%d", &n);
	for(int i = 1; i <= n; i++) {
		scanf("%d", &arr[i].val);
		if(arr[i].val == 1) num1++;
		arr[i].id = i;
	}
	sort(arr + 1, arr + n + 1, cmp);
	if(num1 == n) { printf("NO
"); return 0; }

	for(int i = num1+1; i < n; i++) {
		ans[arr[i].id][arr[i+1].id] = 1;
		cnt++;
		arr[i].val--;
		arr[i+1].val--;
	}
	int p = num1;
	for(int i = n; i > num1; i--) {
		while(arr[i].val && p) {
			cnt++;
			ans[arr[p].id][arr[i].id] = 1;
			arr[i].val--;
			p--;
		}
	}
	if(p) { printf("NO
"); return 0; }
	printf("YES %d
", min(n-num1+1, n-1));
	printf("%d
", cnt);
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++) 
			if(ans[i][j]) printf("%d %d
", i, j);
}

一个a了的答案

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<iostream>
#include<string>
#include<queue>
using namespace std;

int ans[510][510];
struct node{
	int val;
	int id;
}arr[510];

bool cmp(node a, node b) { 
	return a.val < b.val;
}

int main() {
	int n, num1 = 0, cnt = 0;
	scanf("%d", &n);
	for(int i = 1; i <= n; i++) {
		scanf("%d", &arr[i].val);
		if(arr[i].val == 1) num1++;
		arr[i].id = i;
	}
	if(num1 == 0) num1 = 1;
	sort(arr + 1, arr + n + 1, cmp);
	if(num1 == n) { 
		printf("NO
"); 
		return 0; 
	}

	for(int i = num1; i < n; i++) {
		ans[arr[i].id][arr[i+1].id] = 1;
		cnt++;
		arr[i].val--;
		arr[i+1].val--;
	}
	int p = num1-1;
	for(int i = n; i >= num1; i--) {
		while(arr[i].val && p) {
			cnt++;
			ans[arr[p].id][arr[i].id] = 1;
			arr[i].val--;
			p--;
		}
	}
	if(p) { printf("NO
"); return 0; }
	printf("YES %d
", min(n-num1+1, n-1));
	printf("%d
", cnt);
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++) 
			if(ans[i][j]) printf("%d %d
", i, j);
}