Educational Codeforces Round 25

A

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

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#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;
char st[105];
int n;

int main(){
scanf("%d", &n);
scanf("%s", st);
int len = strlen(st);
st[len] = '0';
int i = 0;
while(i <= len){
int cnt = 0;
while(st[i] == '1' && i <= len)	cnt++, ++i;
printf("%d", cnt);
cnt = 0;
while(st[i] == '0' && i <= len)	cnt++, ++i;
for(int i = 0; i < cnt - 1; ++i)
printf("0");
}
puts("");
return 0;
}

B

枚举每一个空位，填上X判断是否有五个连续的X

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

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#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 = 5 + 10;
char mar[qq][qq];
bool Check(){
for(int i = 1; i <= 10; ++i) {
for(int j = 1; j <= 10; ++j) {
int x = i, y = j;
int cnt = 0, num = 0;
while(cnt < 5 && x < 11) {
if(mar[x][y] == 'X')	num++;
cnt++;
x++;
}
if(num == 5)	return true;
cnt = num = 0;
x = i, y = j;
while(cnt < 5 && y < 11) {
if(mar[x][y] == 'X')	num++;
cnt++;
y++;
}
if(num == 5)	return true;
cnt = num = 0;
x = i, y = j;
while(x < 11 && y < 11 && cnt < 5) {
if(mar[x][y] == 'X')	num++;
cnt++, x++, y++;
}
if(num == 5)	return true;
cnt = num = 0;
x = i, y = j;
while(x > 0 && y > 0 && cnt < 5) {
if(mar[x][y] == 'X')	num++;
cnt++, x++, y--;
}
if(num == 5)	return true;
}
}
return false;
}

int main(){
for(int i = 1; i <= 10; ++i) {
scanf("%s", mar[i] + 1);
}
for(int i = 1; i <= 10; ++i) {
for(int j = 1; j <= 10; ++j) {
if(mar[i][j] != '.')	continue;
mar[i][j] = 'X';
if(Check()){
puts("YES");
return 0;
}
mar[i][j] = '.';
}
}
puts("NO");
return 0;
}

C

题意：n道题，已经解决了难度为k的题，能解决某道难度为ai的题的条件是 已经解决过的某道题难度为d， d > ai / 2则可以解决ai这道题，问最少需要在其他OJ解决多少题，才能把给定的
4000
n个题解决。

思路：每次对k进行最大值的取值

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

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#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;
int dif[qq];
int n, k;

int main(){
scanf("%d%d", &n, &k);
for(int i = 1; i <= n; ++i) {
scanf("%d", dif + i);
}
sort(dif + 1, dif + 1 + n);
int cnt = 0;
for(int i = 1; i <= n; ++i) {
if(k * 2 < dif[i]) {
while(k * 2 < dif[i]) {
k = k * 2;
cnt++;
}
k = max(k, dif[i]);
}
k = max(k, dif[i]);
}
printf("%d\n", cnt);
return 0;
}

D

题意：两个字符串s和t，s中由 '?'和小写字母组成，t由小写字母组成，s字符串中可以交换任意两个字符，现在问你填充s中的问号使得s中有最大数量的不相交的t字符串(交换位置后)

思路：- - 可以暴力答案，我这里用的二分

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

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#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 = 1e6 + 10;
char st[qq], t[qq];
LL nums[30], numt[30];
LL num[30], p[30];
bool Check(LL x, LL cnt) {
for(int i = 0; i < 26; ++i) {
if(numt[i] == 0){
num[i] = 0;
continue;
}
if(nums[i] - numt[i] * x >= 0)	num[i] = 0;
else num[i] = numt[i] * x - nums[i];
cnt = cnt - num[i];
if(cnt < 0)	return false;
}
if(cnt < 0)	return false;
return true;
}

int main(){
scanf("%s%s", st, t);
LL cnt = 0;
int lens = strlen(st);
for(int i = 0; i < lens; ++i) {
if(st[i] == '?')	cnt++;
else nums[st[i] - 'a']++;
}
int lent = strlen(t);
for(int i = 0; i < lent; ++i) {
numt[t[i] - 'a']++;
}
LL l = 0, r = 1e9, ans = -1, mid;
while(l <= r) {
mid = (l + r) / 2;
if(Check(mid, cnt)) {
for(int i = 0; i < 26; ++i) {
p[i] = num[i];
}
ans = mid;
l = mid + 1;
}else {
r = mid - 1;
}
}
int j = 0;
for(int i = 0; i < 26; ++i) {
if(p[i] == 0)	continue;
while(p[i] > 0 && j < lens) {
if(st[j] == '?')	st[j] = i + 'a', p[i]--;
j++;
}
}
for(int i = 0; i < lens; ++i) {
if(st[i] == '?')	st[i] = 'a';
}
puts(st);
return 0;
}

E

题意：n个节点，m条有向边，每个节点有一个label值，对于一条有向边u->v， 要求u的label小于v的label，最后按1到n输出每一个节点的label，要求字典序最小

还没看懂这个思路，先挖个坑

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

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#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;
vector<int> vt[qq];
int n, m;
int deg[qq];
int vis[qq];
priority_queue<int> Q;

int main(){
scanf("%d%d", &n, &m);
int a, b;
for(int i = 1; i <= m; ++i) {
scanf("%d%d", &a, &b);
deg[a]++;
vt[b].pb(a);
}
for(int i = 1; i <= n; ++i) {
if(deg[i] == 0)	Q.push(i);
}
int num = n;
while(!Q.empty()) {
int u = Q.top();
Q.pop();
for(int i = 0; i < (int)vt[u].size(); ++i) {
int v = vt[u][i];
deg[v]--;
if(deg[v] == 0)	Q.push(v);
}
vis[u] = num--;
}
for(int i = 1; i <= n; ++i) {
printf("%d ", vis[i]);
}
puts("");
return 0;
}