Codeforces Round #301 (Div. 2) -- （A,B,C,D）

標籤：acm   codeforces   

題目傳送：Codeforces Round #301 (Div. 2)

A. Combination Lock

水題，求最小移動次數，簡單貪心一下即可

AC代碼：

#include <cstdio>#include <cstring>#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 0x7fffffffusing namespace std;int n; char s1[1005];char s2[1005];int main() {scanf("%d", &n);scanf("%s %s", s1, s2);int ans = 0;for(int i = 0; i < n; i ++) {if(s2[i] > s1[i]) {ans += min(s2[i] - s1[i], s1[i] + 10 - s2[i]);}else {ans += min(s1[i] - s2[i], s2[i] + 10 - s1[i]);}}printf("%d\n", ans);return 0;}

B. School Marks

也比較簡單，就是有點繁瑣，具體看代碼吧

AC代碼：

#include <cstdio>#include <cstring>#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 0x7fffffffusing namespace std;int n, k, p, x, y;int a[1005];int main() {scanf("%d %d %d %d %d", &n, &k, &p, &x, &y);int tot = 0;//記錄當前的總和 int cnt = 0;//記錄當前大於y的數目 for(int i = 0; i < k; i ++) {scanf("%d", &a[i]);tot += a[i];if(a[i] >= y) {cnt ++;}}int mid = (n + 1) / 2;int xu;//記錄最少所需要的數 if(k - cnt >= mid) {//當前如果有半數都比y小則輸出-1 printf("-1\n");return 0;}if(cnt < mid) {//比y大的少於mid的情況 xu = (mid - cnt) * y;xu += (n - k - (mid - cnt));if(xu <= x - tot) {for(int i = 0; i < n - k - (mid - cnt); i ++) {printf("1 ");}for(int i = 0; i < mid - cnt; i ++) {printf("%d ", y);}}else {printf("-1\n");}} else {//比y大的大於等於mid的情況 xu = n - k;if(xu <= x - tot) {for(int i = 0; i < n - k; i ++) {printf("1 ");}}else printf("-1\n");}return 0;}

C. Ice Cave

題意：很簡單，就是一個n*m的冰面，有的破碎了，走一次就會陷下去，有的完好的，不過走一次就破碎了，下次走就會陷下去，給你一個起點和終點，看起點能否走到終點那裡陷下去，注意起點肯定是破碎的，且終點可能會和起點相同

思路：首先，特判一下起點和終點相同的情況，然後bfs一下看起點能否能走到終點，然後根據終點的情況分類，當終點為破碎的冰時，只要找到路徑即YES，否則NO，當終點為完好的冰時，到了終點後還要走出去再回來，這裡注意，只要當前挨著的冰有一塊為‘.‘（即完好的），則成立，輸出YES，否則輸出NO；只需要預先處理一下原來終點挨著的冰塊的‘.‘的個數cnt即可（cnt>=2就YES，否則NO），然後這裡需要特判一下起點和終點挨著的情況（因為此時只需要cnt>=1即可，這裡尤其猥瑣）

AC代碼：

#include <cstdio>#include <cstring>#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 0x7fffffffusing namespace std;struct node {int x, y;node(int x,int y) : x(x), y(y) {}};int n, m;int mp[505][505];int mx[4] = {-1, 0, 1, 0};int my[4] = {0, 1, 0, -1};int r1, c1;int r2, c2;char s[505];int bfs() {//bfs找路徑 queue<node> que;mp[r1][c1] --;que.push(node(r1, c1));while(!que.empty()) {node tmp = que.front();que.pop();for(int i = 0; i < 4; i ++) {int xx = tmp.x + mx[i];int yy = tmp.y + my[i];if(xx >= 1 && xx <= n && yy <= m && yy >= 1) {if(xx == r2 && yy == c2) return 1;if(mp[xx][yy] == 2) {mp[xx][yy] --;que.push(node(xx, yy));}}}}return 0;}int main() {scanf("%d %d", &n, &m);for(int i = 0; i < n; i ++) {scanf("%s", s);int len = strlen(s);for(int j = 0; j < len; j ++) {if(s[j] == '.') {mp[i + 1][j + 1] = 2;}else {mp[i + 1][j + 1] = 1;}}}scanf("%d %d %d %d", &r1, &c1, &r2, &c2);int cnt = 0;//記錄終點旁邊有幾個'.' for(int i = 0; i < 4; i ++) {int xx = r2 + mx[i];int yy = c2 + my[i];if(mp[xx][yy] == 2) cnt ++;}if(r1 == r2 && c1 == c2) {//特判一下起點和終點相同的情況 if(cnt >= 1) {printf("YES\n");}else printf("NO\n");return 0;}int flag = 0;//特判一下起點和終點相鄰的情況，這裡特別坑，感覺坑了好多人 for(int i = 0; i < 4; i ++) {int xx = r1 + mx[i];int yy = c1 + my[i];if(xx == r2 && yy == c2) {flag = 1;break;}}if(flag) {if(mp[r2][c2] == 1) {printf("YES\n");}else {if(cnt >= 1) {printf("YES\n");}else printf("NO\n");}return 0;}//起點和終點不相同且不相鄰的情況 if(mp[r2][c2] == 1) {if(bfs()) {printf("YES\n");}else printf("NO\n");}else {if(bfs()) {if(cnt >= 2) {printf("YES\n");}else printf("NO\n");}else printf("NO\n");}return 0;}

D. Bad Luck Island

思路：機率DP，設狀態dp[i][j][k]為此時石頭i個剪刀j個布k個的機率，可以知道dp[i][j][k]肯定由dp[i+1][j][k]，dp[i][j+1][k]，dp[i][j][k+1]得來，初始狀態dp[r][s][p]為1，具體看代碼

AC代碼：

#include <cstdio>#include <cstring>#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 0x7fffffffusing namespace std;int r, s, p;double dp[105][105][105];int main() {cin >> r >> s >> p;dp[r][s][p] = 1;for(int i = r; i >= 0; i --) {for(int j = s; j >= 0; j --) {for(int k = p; k >= 0; k --) {if(i == r && j == s && k == p) continue;double sum = i + j + k + 1; //上一狀態的總數 if(sum <= 1) continue;//全為0的時候就不需要計算了 double t1 = 0, t2 = 0, t3 = 0;double t = 0;t1 = 2.0 * dp[i + 1][j][k] * (i + 1) / sum * k / (sum - 1);//當隨機出現的是相同的人時的機率，這個機率不應該算到dp裡面 t = (i+1)/sum*i/(sum-1) + j/sum*(j-1)/(sum-1) + k/sum*(k-1)/(sum-1);if(t < 1.0) t1 /= (1.0 - t);//通過比例消掉 t2 = 2.0 * dp[i][j + 1][k] * (j + 1) / sum * i / (sum - 1);t = i/sum*(i-1)/(sum-1) + (j+1)/sum*j/(sum-1) + k/sum*(k-1)/(sum-1);if(t < 1.0) t2 /= (1.0 - t);t3 = 2.0 * dp[i][j][k + 1] * (k + 1) / sum * j / (sum - 1);t = i/sum*(i-1)/(sum-1) + j/sum*(j-1)/(sum-1) + (k+1)/sum*k/(sum-1);if(t < 1.0) t3 /= (1.0 - t);dp[i][j][k] = t1 + t2 + t3;//統計上一狀態到目前狀態的機率 }}} double ansr = 0;double anss = 0;double ansp = 0;for(int i = 1; i <= r; i ++) {ansr += dp[i][0][0];}for(int i = 1; i <= s; i ++) {anss += dp[0][i][0];}for(int i = 1; i <= p; i ++) {ansp += dp[0][0][i];}printf("%.12lf %.12lf %.12lf\n", ansr, anss, ansp);return 0;}

Codeforces Round #301 (Div. 2) -- （A,B,C,D）