Codeforces Round #293 (Div. 2)

A. Vitaly and Strings

题意： 判断两个相同长度的字符串字典序之间是否还有其他字符串。

思路： 直接求出s的下个字典序的字符串， 然后判断是不是和t相等。

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#include <stack>#include <string>#include <cmath>#include <set>#include <map>#include <vector>#include <queue>#include <ctime>#include <cstdlib>using namespace std;#define mxn 120#define mxe 60020#define M 2000020#define mod 2147483647#define LL long long#define inf 0x3f3f3f3f#define vi vector<int>#define PB push_back#define MP make_pair#define pii pair<int, int>#pragma comment(linker,"/STACK:1024000000,1024000000")#define ls (i << 1)#define rs (ls | 1)#define md ((ll + rr) >> 1)char s[mxn], t[mxn];void nxt() {int n = strlen(s);for(int i = n - 1; i >= 0; --i) {if(s[i] != 'z') {++s[i];for(int j = i + 1; j < n; ++j)s[j] = 'a';break;}}}int main() {while(cin >> s >> t) {nxt();if(strcmp(s, t) == 0)puts("No such string");elseprintf("%s\n", s);}return 0;}

B. Tanya and Postcard

题意：给定两个字符串s和t， 要求重新排列字符串t， 使得跟s在相同位置上相同字母尽可能多的情况下， 不区分大小写相同的字母尽可能多。

思路： 考虑'A'和‘a'， s里面有多少个’A'， t就要尽量下先用’A'抵消掉， 不够的再用‘a'抵消掉， 最后再用其他字符抵消。

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#include <stack>#include <string>#include <cmath>#include <set>#include <map>#include <vector>#include <queue>#include <ctime>#include <cstdlib>using namespace std;#define mxn 200020#define mxe 60020#define M 2000020#define mod 2147483647#define LL long long#define inf 0x3f3f3f3f#define vi vector<int>#define PB push_back#define MP make_pair#define pii pair<int, int>#pragma comment(linker,"/STACK:1024000000,1024000000")#define ls (i << 1)#define rs (ls | 1)#define md ((ll + rr) >> 1)char s[mxn], t[mxn];int a[130], b[130];int main() {while(scanf("%s", s) != EOF) {memset(a, 0, sizeof a);memset(b, 0, sizeof b);for(int i = 0; s[i]; ++i)a[s[i]]++;scanf("%s", s);for(int i = 0; s[i]; ++i)b[s[i]]++;int x = 0, y = 0;for(int i = 'a'; i <= 'z'; ++i) {int j = i - 'a' + 'A';int c = min(b[i], a[i]);x += c;b[i] -= c, a[i] -= c;int d = min(b[j], a[j]);x += d;b[j] -= d, a[j] -= d;c = a[i] + a[j];y += min(c, b[i] + b[j]);}printf("%d %d\n", x, y);}return 0;}

C. Anya and Smartphone

题意：简单模拟

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#include <stack>#include <string>#include <cmath>#include <set>#include <map>#include <vector>#include <queue>#include <ctime>#include <cstdlib>using namespace std;#define mxn 200020#define mxe 60020#define M 2000020#define mod 2147483647#define LL long long#define inf 0x3f3f3f3f#define vi vector<int>#define PB push_back#define MP make_pair#define pii pair<int, int>#pragma comment(linker,"/STACK:1024000000,1024000000")#define ls (i << 1)#define rs (ls | 1)#define md ((ll + rr) >> 1)int n, m, k;int a[mxn],  id[mxn];int main() {while(scanf("%d%d%d", &n, &m, &k) != EOF) {for(int i = 1; i <= n; ++i) {scanf("%d", &a[i]);id[a[i]] = i;}LL ans = 0;for(int i = 1; i <= m; ++i) {int x;scanf("%d", &x);int t = id[x];ans += (t - 1) / k;++ans;if(id[x] == 1) continue;int val = a[id[x]-1];swap(a[id[x]], a[id[x]-1]);swap(id[x], id[val]);}printf("%I64d\n", ans);}}

D. Ilya and Escalator

题意：有一个n个人的队列和电梯， 每秒钟队首的一个人有p的概率进入电梯， 1-p的概率不进入电梯， 问t秒后电梯里面的期望人数。

解法：dp[i][j] 表示第i天有j个人的概率， dp[i][j] = dp[i-1][j] * p + dp[i-1][j] * (1-p)， 注意一下边界， 然后期望直接求就可以了。

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#include <stack>#include <string>#include <cmath>#include <set>#include <map>#include <vector>#include <queue>#include <ctime>#include <cstdlib>using namespace std;#define mxn 2020#define mxe 60020#define M 2000020#define mod 2147483647#define LL long long#define inf 0x3f3f3f3f#define vi vector<int>#define PB push_back#define MP make_pair#define pii pair<int, int>#pragma comment(linker,"/STACK:1024000000,1024000000")#define ls (i << 1)#define rs (ls | 1)#define md ((ll + rr) >> 1)double dp[mxn][mxn];int n, t;double p;int main() {while(cin >> n >> p >> t) {for(int i = 0; i <= n; ++i)dp[0][i] = 0;dp[0][0] = 1;for(int i = 1; i <= t; ++i) {for(int j = 0; j <= n; ++j) {if(j == 0)dp[i][j] = dp[i-1][j] * (1 - p);else if(j == n)dp[i][j] = dp[i-1][j-1] * p + dp[i-1][j];elsedp[i][j] = dp[i-1][j-1] * p + dp[i-1][j] * (1 - p);}}double ans = 0;for(int i = 0; i <= n; ++i)ans += dp[t][i] * i;printf("%.17lf\n", ans);}return 0;}

E. Arthur and Questions

题意： 给定k和n个数的数列a[]， 其中数列中有一些元素不确定， 问a数组是否存在使得数列a的k连续子序列的和递增， 如果存在， 输出使得a数组绝对值之和最小的方案。

思路：a的k连续子序列的和递增， 也就是a[1] < a[k+1], a[2] < a[k+2], a[3] < a[k+3]...， 把a按照模k分类， 于是变成模k的数里面要连续严格递增，一段一段地考虑不确定连续的数，对一段连续的不确定的数都有一定的限制范围， 假设这段不连续的数最大值必须小于r， 最小值必须大于l， 这段连续的不确定数的长度为len， 如果r - l - 1 < len， 则无解， 如果r <= 0， 这段就尽量向r取， 如果l >= 0， 就尽量向l取， 否则尽量靠0取。

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#include <stack>#include <string>#include <cmath>#include <set>#include <map>#include <vector>#include <queue>#include <ctime>#include <cstdlib>using namespace std;#define mxn 100020#define mxe 60020#define M 2000020#define mod 2147483647#define LL long long#define inf 0x3f3f3f3f#define vi vector<int>#define PB push_back#define MP make_pair#define pii pair<int, int>#pragma comment(linker,"/STACK:1024000000,1024000000")#define ls (i << 1)#define rs (ls | 1)#define md ((ll + rr) >> 1)int a[mxn];int n, k;vector<int> g[mxn];char s[20];int f() {if(s[0] == '?') return inf;bool flag = 0;int i = 0;if(s[0] == '-') {flag = 1;++i;}int ret = 0;for(; s[i]; ++i)ret = ret * 10 + s[i] - '0';if(flag)ret = -ret;return ret;}int q[mxn];bool calc(int i) {int len = g[i].size();int pre = -inf;for(int j = 0; j < len; ++j) {if(g[i][j] == inf) continue;if(g[i][j] <= pre) return 0;pre = g[i][j];}int l = -inf, r = inf;int j = 0;while(j < len) {if(g[i][j] != inf) {l = g[i][j];++j;continue;}int v = j;while(v < len && g[i][v] == inf) ++v;int x = v - j;if(v < len) r = g[i][v];else r = inf;//printf("%d %d\n", l, r);if(r - l - 1 < x) return 0;if(r <= 0) {for(int u = 0; u < x; ++u) {g[i][v-u-1] = r - u - 1;}}else if(l >= 0) {for(int u = 0; u < x; ++u)g[i][j+u] = l + u + 1;}else {int head = 0, tail = 0;q[tail++] = 0;for(int i = 1; ; ++i) {if(tail == x) break;if(i < r) {q[tail++] = i;}if(tail == x) break;if(-i > l) {q[tail++] = -i;}}sort(q, q + tail);for(int u = 0; u < tail; ++u)g[i][u+j] = q[u];}l = g[i][v];j = v + 1;}return 1;}int main() {//freopen("tt.txt", "r", stdin);while(scanf("%d%d", &n, &k) != EOF) {for(int i = 0; i < n; ++i) {scanf("%s", s);g[i%k].PB(f());}bool flag = 1;for(int i = 0; i < k; ++i)if(!calc(i)) {flag = 0;break;}if(!flag)puts("Incorrect sequence");elsefor(int i = 0; i < n; ++i)printf("%d%c", g[i%k][i/k], i == n - 1? '\n': ' ');}return 0;}

F. Pasha and Pipe

题意：给定一个n*m的字符矩阵， * 表示 不可行， .表示可行， 现在要求放一条水管， 满足n多要求， 问方案数。

思路：

分析要求后可以发现， 放水管方案有4类：

对于第1,2种， 将矩形旋转4次可以计算出所有情况， 

对于第3,4种， 将矩形旋转2次可以计算出所有情况，

其中， 对于第二种情况， 两条横边行坐标之差要大于1， 第三种情况包含了第四种， 算的时候要减去， 第三种和第四种不能一起算， 因为列数等于2的时候第三种情况不包含第四种。

第一种情况： O(nm)枚举转折点， 然后判断一个点的上边和左边是不是全都是'.'；

第二种情况：枚举转折点所在的列， 然后对每个列， 考虑一段连续的'.'， 假设有t行左边全部都是'.'，则答案加上C(t, 2)， 在减去连续的行。

第三种情况：类似第二种。

第四种情况：O(n)直接求。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <queue>#include <stack>#include <set>#include <map>#include <cstdlib>using namespace std;#define LL long long#define inf 0x3f3f3f3f#define eps 1e-8#define ULL unsigned long long#define mxn 2020#define mxe 200020#define mxnode 20020#define mod 1000000007#define vi vector<int>#define pii pair<int, int>#define vii vector<pii >char g[mxn][mxn];char t[mxn][mxn];int cnt[mxn][mxn][2];int n, m;void rt() {for(int i = 1; i <= n; ++i)for(int j = 1; j <= m; ++j)t[m-j+1][i] = g[i][j];swap(n, m);memcpy(g, t, sizeof g);}void getCnt() {memset(cnt, 0, sizeof cnt);for(int i = 1; i <= n; ++i) {for(int j = 1; j <= m; ++j) {int c = g[i][j] == '.';cnt[i][j][0] = cnt[i-1][j][0] + c;cnt[i][j][1] = cnt[i][j-1][1] + c;}}}LL f1() {LL ret = 0;memset(cnt, 0, sizeof cnt);getCnt();for(int i = 2; i < n; ++i) {for(int j = 2; j < m; ++j) {if(cnt[i][j][0] == i && cnt[i][j][1] == j)++ret;}}return ret;}LL f2() {LL ret = 0;getCnt();for(int j = 2; j < m; ++j) {int i = 2;while(i < n) {while(i < n && g[i][j] == '*') ++i;if(i >= n) break;int r = i;while(r < n && g[r][j] == '.') ++r;--r;int t = 0;for(int x = i; x <= r; ++x)if(cnt[x][j][1] == j)++t;if(t > 0)ret += t * (t - 1) / 2;for(int x = i; x < r; ++x) {if(cnt[x][j][1] == j && cnt[x+1][j][1] == j)--ret;}i = r + 2;}}return ret;}LL f3() {LL ret = 0;getCnt();for(int j = 2; j < m; ++j) {int i = 2;while(i < n) {while(i < n && g[i][j] == '*') ++i;if(i >= n) break;int r = i;while(r < n && g[r][j] == '.') ++r;--r;int t = 0;for(int x = i; x <= r; ++x)if(cnt[x][j][1] == j)++t;for(int x = i; x <= r; ++x) {if(cnt[x][m][1] - cnt[x][j][1] == m - j) {ret += t;if(cnt[x][j][1] == j)--ret;}}i = r + 2;}}return ret;}LL f4() {LL ret = 0;getCnt();for(int i = 2; i < n; ++i) {if(cnt[i][m][1] == m)++ret;}return ret;}void out() {for(int i = 1; i <= n; ++i) {for(int j = 1; j <= m; ++j)putchar(g[i][j]);puts("");}}int main() {//freopen("tt.txt", "r", stdin);while(scanf("%d%d", &n, &m) != EOF) {for(int i = 1; i <= n; ++i)scanf("%s", g[i] + 1);LL ans = 0;for(int i = 0; i < 4; ++i) {ans += f1();rt();//out();}//printf("%I64d\n", ans);for(int i = 0; i < 4; ++i) {ans += f2();rt();}//printf("%I64d\n", ans);for(int i = 0; i < 2; ++i) {ans += f3();rt();}//printf("%I64d\n", ans);for(int i = 0; i < 2; ++i) {ans += f4();rt();}printf("%I64d\n", ans);}return 0;}