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水?A. A and B and Chess

/*水題 */ #include <cstdio> #include <algorithm> #include <iostream> #include <cstring> #include <string> using namespace std;const int maxn = 1e6 + 10; int a[maxn];int main(void) {//freopen ("A.in", "r", stdin);string s1;int suma = 0; int sumb = 0;for (int i=1; i<=8; ++i){cin >> s1;for (int j=0; s1[j]!='\0'; ++j){if (s1[j] == '.') continue;else if (s1[j] == 'Q') suma += 9;else if (s1[j] == 'R') suma += 5;else if (s1[j] == 'B') suma += 3;else if (s1[j] == 'N') suma += 3;else if (s1[j] == 'P') suma += 1;else if (s1[j] == 'q') sumb += 9;else if (s1[j] == 'r') sumb += 5;else if (s1[j] == 'b') sumb += 3;else if (s1[j] == 'n') sumb += 3;else if (s1[j] == 'p') sumb += 1;}}if (suma > sumb) cout << "White" << endl;else if (suma < sumb) cout << "Black" << endl;else cout << "Draw" << endl;return 0; }

水?B. A and B and Compilation Errors

題意：三組數(shù)列，依次少一個(gè)，找出少了的兩個(gè)數(shù)

思路：

1. 三次排序，逐個(gè)對(duì)比（如果沒找到，那個(gè)數(shù)在上一個(gè)數(shù)列的末尾）
2. 求和做差，最簡(jiǎn)單！

#include <cstdio> #include <algorithm> #include <iostream> using namespace std;const int maxn = 1e5 + 10; int a[maxn]; int b[maxn]; int c[maxn];int main(void) {//freopen ("B.in", "r", stdin);int n, x, y;while (~scanf ("%d", &n)){x = y = 0;for (int i=1; i<=n; ++i){scanf ("%d", &a[i]);}sort (a+1, a+1+n);for (int i=1; i<=n-1; ++i){scanf ("%d", &b[i]);}sort (b+1, b+1+n-1);for (int i=1; i<=n-1; ++i){if (a[i] == b[i]) continue;else{x = a[i];break;}}if (x == 0) x = a[n];for (int i=1; i<=n-2; ++i){scanf ("%d", &c[i]);}sort (c+1, c+1+n-2);for (int i=1; i<=n-2; ++i){if (b[i] == c[i]) continue;else{y = b[i];break;}}if (y == 0) y = b[n-1];printf ("%d\n%d\n", x, y);}return 0; }/* #include <cstdio> #include <algorithm> #include <iostream> using namespace std;const int maxn = 1e5 + 10; int a[maxn]; int b[maxn]; int c[maxn]; int suma, sumb, sumc;int main(void) {//freopen ("B.in", "r", stdin);int n;while (~scanf ("%d", &n)){suma = sumb = sumc = 0;for (int i=1; i<=n; ++i){scanf ("%d", &a[i]); suma += a[i];}for (int i=1; i<=n-1; ++i){scanf ("%d", &b[i]); sumb += b[i];}for (int i=1; i<=n-2; ++i){scanf ("%d", &c[i]); sumc += c[i];}printf ("%d\n%d\n", suma - sumb, sumb - sumc);}return 0; } */

構(gòu)造?C. A and B and Team Training

題意：方案：高手1和菜鳥2 或者 高手2菜鳥1 三人組隊(duì)求最大組隊(duì)數(shù)
思路：

1. 高手加菜鳥每三個(gè)分開，在n，m的數(shù)字之內(nèi)
2. 高手多，高手2；菜鳥多，菜鳥2 比較好理解

#include <cstdio> #include <algorithm> using namespace std;int main(void) {//freopen ("C.in", "r", stdin);int n, m;while (~scanf ("%d%d", &n, &m)){int ans = (n + m) / 3;ans = min (ans, n);ans = min (ans, m);printf ("%d\n", ans);}return 0; }/* #include <cstdio> #include <algorithm> using namespace std;int main(void) {//freopen ("C.in", "r", stdin);int n, m;while (~scanf ("%d%d", &n, &m)){int cnt = 0;while (n && m && (n + m) >= 3){if (n >= m){n -= 2; m -= 1;}else{n -=1; m -= 2;}cnt++;}printf ("%d\n", cnt);return 0; } */

　　

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轉(zhuǎn)載于:https://www.cnblogs.com/Running-Time/p/4366782.html

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