Codeforces Round #369 (Div. 2) ABCDE题解

A. Bus to Udayland

水题，枚举字符并判断，无坑点

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef __int64 LL;
typedef pair<int,int> PII;
#define FIN freopen("in.txt", "r", stdin);
#define FOUT freopen("out.txt", "w", stdout);
#define lson l, mid, cur << 1
#define rson mid + 1, r, cur << 1 | 1
//#pragma comment(linker, "/STACK:1024000000,1024000000")
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const double ERR = 1e-8;
const int MOD = 1e9 + 7;
const int MAXN = 1e3 + 50;
const int MAXM = 5e3 + 50;

int n, m;
char s[MAXN][10];

int main() {
#ifdef LOCAL_NORTH
FIN;
#endif // LOCAL_NORTH
while (~scanf("%d", &n)) {
bool ok = false;
for (int i = 0; i < n; i++) {
scanf("%s", s[i]);
if (!ok) {
if (s[i][0] == 'O' && s[i][1] == 'O') {
s[i][0] = s[i][1] = '+';
ok = true;
}
if (!ok) {
if (s[i][3] == 'O' && s[i][4] == 'O') {
s[i][3] = s[i][4] = '+';
ok = true;
}
}
}
}
if (!ok)
puts("NO");
else {
puts("YES");
for (int i = 0; i < n; i++)
printf("%s\n", s[i]);
}
}
#ifdef LOCAL_NORTH
cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC * 1000 << " ms." << endl;
#endif // LOCAL_NORTH
return 0;
}

B. Chris and Magic Square
水题，先记录0出现的位置，然后用已知的行和，求出应该填写的数，再判断是否相等。注意要特判n=1的情况。

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef __int64 LL;
typedef pair<int,int> PII;
#define FIN freopen("in.txt", "r", stdin);
#define FOUT freopen("out.txt", "w", stdout);
#define lson l, mid, cur << 1
#define rson mid + 1, r, cur << 1 | 1
//#pragma comment(linker, "/STACK:1024000000,1024000000")
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const double ERR = 1e-8;
const int MOD = 1e9 + 7;
const int MAXN = 500 + 50;
const int MAXM = 5e3 + 50;

int n, num[MAXN][MAXN];
LL c[MAXN], r[MAXN], d1, d2;

int main() {
#ifdef LOCAL_NORTH
FIN;
#endif // LOCAL_NORTH
while (~scanf("%d", &n)) {
memset(c, 0, sizeof(c));
memset(r, 0, sizeof(r));
d1 = d2 = 0;
int x, y;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%d", &num[i][j]);
if (num[i][j] == 0) {
x = i; y = j;
continue;
}
r[i] += num[i][j];
c[j] += num[i][j];
if (i == j) d1 += num[i][j];
if (i + j == n - 1) d2 += num[i][j];
}
}
if (n == 1) {
puts("1");
continue;
}
LL ans = 0;
if (x > 0)
ans = r[x - 1] - r[x];
else
ans = r[x + 1] - r[x];
r[x] += ans;
c[y] += ans;
if (x == y)
d1 += ans;
if (x + y == n - 1)
d2 += ans;
bool ok = true;
LL flag = r[0];
for (int i = 0; i < n; i++) {
if (r[i] != flag || c[i] != flag) {
ok = false;
break;
}
}
if (d1 != flag || d2 != flag)
ok = false;
if (!ok || ans <= 0)
puts("-1");
else
printf("%I64d\n", ans);
}
#ifdef LOCAL_NORTH
cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC * 1000 << " ms." << endl;
#endif // LOCAL_NORTH
return 0;
}

C. Coloring Trees

简单dp，dp[i][j][k]表示第i个位置，颜色为j，段数为k的最小花费。我比较蠢- -，分了4种情况（当前位置和上一个位置是否为0）讨论（比赛时有个细节写错了结果fst。状态转移方程见代码。时间复杂度是O(n*m^2*k)，用cf的测评机跑不会超时。

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef __int64 LL;
typedef pair<int,int> PII;
#define FIN freopen("in.txt", "r", stdin);
#define FOUT freopen("out.txt", "w", stdout);
#define lson l, mid, cur << 1
#define rson mid + 1, r, cur << 1 | 1
//#pragma comment(linker, "/STACK:1024000000,1024000000")
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const double ERR = 1e-8;
const int MOD = 1e9 + 7;
const int MAXN = 100 + 50;
const int MAXM = 5e3 + 50;

int n, m, k, c[MAXN], cost[MAXN][MAXN];
LL dp[MAXN][MAXN][MAXN];

int main() {
#ifdef LOCAL_NORTH
FIN;
#endif // LOCAL_NORTH
while (~scanf("%d%d%d", &n, &m, &k)) {
memset(dp, INFLL, sizeof(dp));
memset(cost, 0, sizeof(cost));
for (int i = 1; i <= n; i++)
scanf("%d", &c[i]);
if (!c[1])
c[0] = 101;
else
c[0] = c[1];
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
scanf("%d", &cost[i][j]);
}
}
for (int i = 1; i <= m; i++) {
dp[0][i][0] = 0;
}
for (int i = 1; i <= m; i++) {
if (c[1])
dp[1][c[1]][1] = 0;
else
dp[1][i][1] = cost[1][i];
}
for (int i = 1; i <= n; i++) {
if (c[i]) {
if (c[i - 1]) {
for (int l = 1; l < i; l++) {
if (c[i] == c[i - 1])
dp[i][c[i]][l] = min(dp[i][c[i]][l], dp[i - 1][c[i - 1]][l]);
else
dp[i][c[i]][l + 1] = min(dp[i][c[i]][l + 1], dp[i - 1][c[i - 1]][l]);
}
} else {
for (int j = 1; j <= m; j++) {
for (int l = 1; l < i; l++) {
if (c[i] == j)
dp[i][j][l] = min(dp[i][j][l], dp[i - 1][j][l]);
else
dp[i][c[i]][l + 1] = min(dp[i][c[i]][l + 1], dp[i - 1][j][l]);
}
}
}
} else {
if (c[i - 1]) {
for (int j = 1; j <= m; j++) {
for (int l = 1; l < i; l++) {
if (j == c[i - 1])
dp[i][j][l] = min(dp[i][j][l], dp[i - 1][j][l] + cost[i][j]);
else
dp[i][j][l + 1] = min(dp[i][j][l + 1], dp[i - 1][c[i - 1]][l] + cost[i][j]);
}
}
} else {
for (int j1 = 1; j1 <= m; j1++) {
for (int j2 = 1; j2 <= m; j2++) {
for (int l = 1; l < i; l++) {
if (j1 == j2)
dp[i][j1][l] = min(dp[i][j1][l], dp[i - 1][j1][l] + cost[i][j1]);
else
dp[i][j1][l + 1] = min(dp[i][j1][l + 1], dp[i - 1][j2][l] + cost[i][j1]);
}
}
}
}
}
}
LL ans = INFLL;
if (c
)
ans = min(ans, dp
[c
][k]);
else
for (int i = 1; i <= m; i++) {
ans = min(ans, dp
[i][k]);
}
printf("%I64d\n", ans == INFLL ? -1 : ans);
}
#ifdef LOCAL_NORTH
cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC * 1000 << " ms." << endl;
#endif // LOCAL_NORTH
return 0;
}

D. Directed Roads

简单图论+计数。（qwb说的对，确实是水题
题目意思是给你一个可能有环的有向图，你可以翻转一些边（原来u->v变成v->u），使得原图变为一个DAG，输出方案数。
思路比较直接，跑一遍tarjan统计出每个强连通分量的节点数，对于节点数num>2的强连通分量来说，我们必须翻转其中的[1,num-1]（也就是（2^num）-2）条边。然后对于其余强连通分量（也就是不属于num>2的scc的节点），我们可选可不选，也就是有2^tmp种情况。然后把以上两种情况的所有结果相乘即可。

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef __int64 LL;
typedef pair<int,int> PII;
#define FIN freopen("in.txt", "r", stdin);
#define FOUT freopen("out.txt", "w", stdout);
#define lson l, mid, cur << 1
#define rson mid + 1, r, cur << 1 | 1
//#pragma comment(linker, "/STACK:1024000000,1024000000")
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const double ERR = 1e-8;
const int MOD = 1e9 + 7;
const int MAXN = 2e5 + 50;
const int MAXM = 2e5 + 50;

int n;
struct edge {
int to, nxt;
} E[MAXM];
int Head[MAXN], tot;
void edge_init() {
tot = 0;
memset(Head, -1, sizeof(Head));
}
void edge_add(int u, int v) {
E[tot].to = v;
E[tot].nxt = Head[u];
Head[u] = tot++;
}

/*
for (int i = 1; i <= n; i++)
if (!DFN[i])
Tarjan(i);
*/
int Low[MAXN], DFN[MAXN], Stack[MAXN], Belong[MAXN], num[MAXN];
int Index, top, scc;
bool InStack[MAXN];
void Tarjan_init() {
memset(DFN, 0, sizeof(DFN));
memset(num, 0, sizeof(num));
memset(InStack, false, sizeof(InStack));
Index = scc = top = 0;
}
void Tarjan(int u) {
int v;
Low[u] = DFN[u] = ++Index;
Stack[top++] = u;
InStack[u] = true;
for(int i = Head[u]; i != -1; i = E[i].nxt) {
v = E[i].to;
if( !DFN[v] ) {
Tarjan(v);
if( Low[u] > Low[v] )Low[u] = Low[v];
} else if(InStack[v] && Low[u] > DFN[v])
Low[u] = DFN[v];
}
if(Low[u] == DFN[u]) {
scc++;
do {
v = Stack[--top];
InStack[v] = false;
Belong[v] = scc;
num[scc]++;
} while( v != u);
}
}

LL fastpow(LL a, LL c) {
LL res = 1;
while (c) {
if (c & 1)
res = (res * a) % MOD;
a = (a * a) % MOD;
c >>= 1;
}
return res;
}

int main() {
#ifdef LOCAL_NORTH
FIN;
#endif // LOCAL_NORTH
while (~scanf("%d", &n)) {
edge_init();
Tarjan_init();
for (int i = 1; i <= n; i++) {
int t;
scanf("%d", &t);
edge_add(i, t);
}
for (int i = 1; i <= n; i++)
if (!DFN[i])
Tarjan(i);
LL ans = 1;
int tmp = n;
for (int i = 1; i <= scc; i++) {
if (num[i] > 1) {
ans = (ans * (fastpow(2, num[i]) - 2)) % MOD;
tmp -= num[i];
}
}
if (tmp)
ans = (ans * fastpow(2, tmp)) % MOD;
printf("%I64d\n", ans);
}
#ifdef LOCAL_NORTH
cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC * 1000 << " ms." << endl;
#endif // LOCAL_NORTH
return 0;
}

E. ZS and The Birthday Paradox

简单数学题。
一年2^n天，求k个人中有人生日是同一天的概率（1<=n,k<=1e18）。
假如一年365天，n个人中有人生日相同的概率为



所以这一题就是让求



然后求出分子分母的gcd约分就可以了。gcd一定是2^x的形式，因为分母已经是2^x的形式，所以只需要求出分子的因子2的个数。
对于1*2*3*...*k，因子2的个数为k/2+k/4+k/8...。
然后求出gcd的逆元，分子分母同时乘上逆元就得到结果了。

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef __int64 LL;
typedef pair<int,int> PII;
#define FIN freopen("in.txt", "r", stdin);
#define FOUT freopen("out.txt", "w", stdout);
#define lson l, mid, cur << 1
#define rson mid + 1, r, cur << 1 | 1
#define bitcnt(x) __builtin_popcount(x)
#define bitcntll(x) __builtin_popcountll(x)
//#pragma comment(linker, "/STACK:1024000000,1024000000")
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const double ERR = 1e-8;
const int MOD = 1e6 + 3;
const int MAXN = 2e5 + 50;
const int MAXM = 2e5 + 50;

LL n, k;

LL fastpow(LL a, LL c) {
LL res = 1;
while (c) {
if (c & 1)
res = (res * a) % MOD;
a = (a * a) % MOD;
c >>= 1;
}
return res;
}

int main() {
#ifdef LOCAL_NORTH
FIN;
#endif // LOCAL_NORTH
while (~scanf("%I64d%I64d", &n, &k)) {
if (n <= 62 && k > (1LL << n)) {
printf("1 1\n");
continue;
}
LL cnt = 0;
LL p = fastpow(2, n);
for (LL i = k - 1; i >= 1; i >>= 1) {
cnt += i / 2;
}
//        for (LL i = tmp - 1; i >= tmp - k + 1; i--) {
//            LL ttmp = i & (-i);
//            cnt += (LL)log2(ttmp);
//        }
LL inv = fastpow(fastpow(2, cnt), MOD - 2);
LL up = 1, down = fastpow(p, k - 1);
for (LL i = 1; i <= k - 1; i++) {
LL tmp = (p - i + MOD) % MOD;
up = (up * tmp) % MOD;
if (tmp <= 0) //！！！
break;
}
up = up * inv % MOD;
down = down * inv % MOD;
up = (down - up + MOD) % MOD;
printf("%I64d %I64d\n", up, down);
}
#ifdef LOCAL_NORTH
cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC * 1000 << " ms." << endl;
#endif // LOCAL_NORTH
return 0;
}