HDU 4647 Another Graph Game
如果没有边的作用，显然轮流拿当前的最大值即可。
加上边的作用，将边权平均分给两个点，如果一个人选走一条边的两个点，就获得了边的权值；如果分别被两个人拿走，两人的差值不变。
将边权平均分配给点，对点的权值排序轮流选择。
1 #include<cstdio> 2 #include<algorithm> 3 #include<iostream> 4 #define MAXN 100010 5 #define EPS 1e-8 6 typedef long long LL; 7 using namespace std; 8 double arr[MAXN]; 9 int main() { 10 int n, m; 11 int i; 12 int x, y, val; 13 double a, b; 14 while (~scanf("%d%d", &n, &m)) { 15 for (i = 1; i <= n; i++) { 16 scanf("%lf", &arr[i]); 17 } 18 for (i = 0; i < m; i++) { 19 scanf("%d%d%d", &x, &y, &val); 20 arr[x] += val * 0.5 + EPS; 21 arr[y] += val * 0.5 + EPS; 22 } 23 sort(arr + 1, arr + 1 + n); 24 a = b = 0; 25 for (i = n; i > 0; i -= 2) { 26 a += arr[i]; 27 b += arr[i - 1]; 28 } 29 cout << (LL) (a - b + EPS) << endl; 30 } 31 return 0; 32 }
HDU 4648 Magic Pen 6
求前缀和，枚举擦除的起点，求出最右端的位置，更新答案。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define MAXN 100010 5 using namespace std; 6 int arr[MAXN]; 7 int sum[MAXN]; 8 int pos[MAXN]; 9 int main() { 10 int n, m; 11 int i; 12 int ans; 13 while (~scanf("%d%d", &n, &m)) { 14 sum[0] = 0; 15 memset(pos, -1, sizeof(pos)); 16 for (i = 1; i <= n; i++) { 17 scanf("%d", &arr[i]); 18 arr[i] = (arr[i] % m + m) % m; 19 sum[i] = (sum[i - 1] + arr[i]) % m; 20 pos[sum[i]] = i; 21 } 22 ans = 0; 23 for (i = 1; i <= n; i++) { 24 ans = max(ans, pos[sum[i - 1]] - i + 1); 25 } 26 printf("%d\n", ans); 27 } 28 return 0; 29 }
HDU 4649 Professor Tian
dp[i][j][k]表示第i位，处理到第j个数，该位的结果为k的概率。
答案就是每一位的dp[i][n][1]*(1<<i)相加。
1 #include<cstdio> 2 #include<cstring> 3 #define MAXN 220 4 #define MAXM 20 5 double dp[MAXM][MAXN][2]; 6 int arr[MAXN]; 7 char str[MAXN]; 8 double p[MAXN]; 9 int main() { 10 int n; 11 int ca = 1; 12 int i, j, k, l; 13 int tmp; 14 double ans; 15 while (~scanf("%d", &n)) { 16 for (i = 0; i <= n; i++) { 17 scanf("%d", &arr[i]); 18 } 19 for (i = 1; i <= n; i++) { 20 scanf(" %c", &str[i]); 21 } 22 for (i = 1; i <= n; i++) { 23 scanf("%lf", &p[i]); 24 } 25 memset(dp, 0, sizeof(dp)); 26 for (i = 0; i < MAXM; i++) { 27 if (arr[0] & (1 << i)) { 28 dp[i][0][1] = 1; 29 } else { 30 dp[i][0][0] = 1; 31 } 32 for (j = 1; j <= n; j++) { 33 if (arr[j] & (1 << i)) { 34 l = 1; 35 } else { 36 l = 0; 37 } 38 for (k = 0; k < 2; k++) { 39 if (str[j] == '&') { 40 tmp = k & l; 41 } else if (str[j] == '|') { 42 tmp = k | l; 43 } else { 44 tmp = k ^ l; 45 } 46 dp[i][j][tmp] += dp[i][j - 1][k] * (1 - p[j]); 47 dp[i][j][k] += dp[i][j - 1][k] * p[j]; 48 } 49 } 50 } 51 ans = 0; 52 for (i = 0; i < MAXM; i++) { 53 ans += dp[i][n][1] * (1 << i); 54 } 55 printf("Case %d:\n%lf\n", ca++, ans); 56 } 57 return 0; 58 }
