USACO 3.2 分析

 題目一：Factorials

題目大意：求n！的最後非0位。

算法：模擬

1.2的因子的個數顯然比5多很多；

2.可以預先根據5的個數估計出末尾0的個數，然後只需保存多餘2的個數，去更新末位的值；

3.實際做的時候沒有想到上述方法，而是鬼使神差地想到保存末5位，但是無法證明！

 #include <stdio.h> #include <string.h> unsigned int s[32][32]; void dfs(unsigned int m, int leftone, int dep) { //printf("%d %d %d/n", m, leftone, dep); if (dep == 0) { printf("/n"); return; } if (m <= s[dep-1][leftone]) { printf("0"); dfs(m, leftone, dep-1); } else { printf("1"); dfs(m-s[dep-1][leftone], leftone-1, dep-1); } } int main(){ freopen("kimbits.in", "r", stdin); freopen("kimbits.out", "w", stdout); int n, k, m, i, j; scanf("%d %d %u", &n, &k, &m);//%u used for reading unsigned 10-based numbers memset(s,0,sizeof(s)); for (j = 0; j <= k; j++) s[0][j] = 1; for (i = 1; i <= n; i++) { s[i][0] = 1; for (j = 1; j <= k; j++) s[i][j] = s[i-1][j-1] + s[i-1][j]; } dfs(m, k, n); return 0; }

 

題目二：Stringsobits

題目大意：在所有長度爲N的1的位數不超過K的二進制數中（包含前導0），求出第I大的數

算法：DP

狀態轉移方程：S[I][J]表示長度爲I用1的個數不大於J的二進制數的個數；

S[I][J] = S[I-1][J]+S[I-1][J-1]，很像C(A,B)的遞推公式啊！

初始條件：s[0][i] = 1,1 <= i <= K

還有一點要注意的是2^31 超過LONGINT的表示範圍，可以用UNSIGHED LONGINT，讀入格式scanf("%u", &a)

#include <stdio.h> #include <string.h> int main(){ freopen("fact4.in", "r", stdin); freopen("fact4.out", "w", stdout); int n; scanf("%d", &n); int s = 1; for (int i = 1; i <= n; i++) if (i % 5 != 0) s = (s*i) % 100000; else { int j = i; while (j % 5 == 0) s >>= 1, j /= 5; s = (s*j) % 100000; } printf("%d/n", s%10); return 0; }

 

題目三：Spinning Wheels

題目大意：有若干個有缺口的轉動着的輪子，轉的過程是離散的，以時間1爲基本單位，求出什麼時候在某處光線可以通過。

一種思路是狀態仍是離散保存，只保存起始點和末端點，用時間更新初始座標，判斷的時候考慮到有些段可能跨越360的臨界，就要分別判斷是在360內還是360外。

特別注意：只要有一點縫隙，光就可以過！

 #include <stdio.h> #include <string.h> int speed[5], t[5]; int a[5][5], b[5][5]; int main(){ freopen("spin.in", "r", stdin); freopen("spin.out", "w", stdout); int i, j, k; for (i = 0; i < 5; i++) { scanf("%d %d", &speed[i], &t[i]); for (j = 0; j < t[i]; j++) { scanf("%d %d", &a[i][j], &b[i][j]); b[i][j] += a[i][j]; b[i][j] %= 360; } } int tt = 0; bool find = false; do { //CHECK for (i = 0; i < 360 && !find; i++) { bool go = true; for (j = 0; j < 5; j++) { bool flag = false;//if any wedge can through it then flag = true for (k = 0; k < t[j] && !flag; k++) { if (b[j][k] < a[j][k]) b[j][k] += 360; if (a[j][k] <= i && i <= b[j][k] || a[j][k] <= i+360 && i+360 <= b[j][k]) flag = true; //the light between also go through [1,2],[2,3],two ways it go through b[j][k] %= 360; } if (!flag) { go = false; break; } } if (go) { find = true; break; } } if (find) break; //Rotate tt++; for (i = 0; i < 5; i++) for (j = 0; j < t[i]; j++) { a[i][j] = (a[i][j] + speed[i]) % 360; b[i][j] = (b[i][j] + speed[i]) % 360; } } while(tt <= 360); if (find) printf("%d/n", tt); else printf("none/n"); return 0; }

 

還有一種思路是用BOOL來保存狀態，效率相對較低；

 

題目四：Feed Ratios

1.由於數據比較小，可以BRUTE FORCE，關鍵在於3點共線判斷的方法，不推薦用斜率的方法去做，最好用向量的方法去做；

#include <stdio.h> #include <string.h> int s[4][4], get[4]; bool check(int i, int j) { if (s[0][i]*get[j]-s[0][j]*get[i] == 0) return true; else return false; } int main(){ freopen("ratios.in", "r", stdin); freopen("ratios.out", "w", stdout); for (int i = 0; i <= 3; i++) for (int j = 1; j <= 3; j++) scanf("%d", &s[i][j]); int tot = 300, totx, toty, totz, t, tt; for (int i = 0; i < 100 && i < tot; i++) for (int j = 0; j < 100 && i+j < tot; j++) for (int k = 0; k < 100 && i+j+k < tot && i+j+k>0; k++) { for (int l = 1; l <= 3; l++) get[l] = i*s[1][l] + j*s[2][l] + k*s[3][l]; tt = 0; for (int l = 1; l <= 3; l++) if (get[l] != 0 && s[0][l] != 0) { tt = get[l] / s[0][l]; break; } bool flag = true; for (int l = 1; l <= 3 && flag; l++) if (tt * s[0][l] != get[l]) flag = false; /* if (tt > 0 && check(1,2) && check(2,3) && check(3,1)) {// 也可用行列式去做 */ if (flag) { tot = i+j+k; totx = i; toty = j; totz = k; t = tt; } } if (tot == 300) printf("NONE/n"); else printf("%d %d %d %d/n", totx, toty, totz, t); return 0; }

2.本題可以用Cream方程組的知識去做，以下是USACO官方的解答When you combine multiples of mixtures, you can look at it as a multiplication of a matrix by a vector. For example, 8*(1:2:3) + 1*(3:7:1) + 5*(2:1:2) = (21:28:35) = 7*(3:4:5) can be seen as [ 1 3 2 ] [ 8 ] [ 2 7 1 ] * [ 1 ] = 7 [3 4 5] [ 3 1 2 ] [ 5 ] The matrix and the goal ratio vector (3:4:5 in this case) are given; what we have to find is the multiple vector (8:1:5 in this case) and the proportionality costant (7 here). This is like solving a system of linear equations. We can write it as AX = kB. Now, if we use Cramer's Rule, and let D = determinant of A, then X_1 = k D_1 / D X_2 = k D_2 / D X_3 = k D_3 / D, where D_1 is the determinant of the matrix A with the 1st column is replaced by B, D_2 is the determinant of the matrix A with the 2nd column is replaced by B, D_3 is the determinant of the matrix A with the 3rd column is replaced by B. (see a Linear Algebra textbook on why this works.) ,P> We are looking for integral solutions. If D = 0, no solutions. Otherwise, let k = D, and then X_1 = D_1, etc. If these values (X_1,X_2,X_3, _and_ k) all have a greatest common factor above 1, divide them all by that factor, as we are looking for the smallest possible solutions. Now, if some of the numbers is greater than 100, we have not found a feasible solution, so we output `NONE'. Otherwise, the triple (X_1,X_2,X_3) is output, as well as the value k.

 

題目五：Magic Squares

算法：BFS

1.康託展開（Cantor），給出一個1-N的排列，求1-N的全排列字典序中的位置；

2.康託展開的逆運算是給出位置，找到這個排列。

3.這道題要用Cantor + Hash可以有效節約空間

#include <stdio.h> #include <string.h> const int d[3][8] = {{7,6,5,4,3,2,1,0}, {3,0,1,2,5,6,7,4}, {0,6,1,3,4,2,5,7}}, max = 40321; int fra[8]; int q[max][8], best[max], choose[max], pre[max], path[100]; bool vis[max], cov[9]; int head = 0, tail = 0, now[8]; int hash(int *a) { int num = 0; memset(cov,0,sizeof(cov)); for (int i = 0; i < 8; i++) { for (int j = 1; j < a[i]; j++) if (!cov[j]) num += fra[7-i]; cov[a[i]] = 1; } return num; } void check() { int num = hash(now); if (!vis[num]) { tail++; best[tail] = best[head]+1; vis[num] = 1; } } int main(){ freopen("msquare.in", "r", stdin); freopen("msquare.out", "w", stdout); fra[0] = 1; for (int i = 1; i <= 7; i++) fra[i] = fra[i-1]*i; for (int i = 0; i <= 7; i++) scanf("%d", &q[0][i]); best[0] = 0; memset(vis,0,sizeof(vis)); vis[0] = 1; pre[0] = -1; int final = hash(q[0]); for (int i = 0 ; i <= 7; i++) q[0][i] = i+1; //printf("%d/n", final); if (final == 0) printf("0/n/n"); else//why？？？ while (head <= tail) { for (int i = 0; i < 3; i++) { for (int j = 0; j < 8; j++) now[j] = q[head][d[i][j]]; int num = hash(now); if (!vis[num]) { tail++; vis[num] = true; for (int j = 0; j < 8; j++) q[tail][j] = now[j]; best[tail] = best[head] + 1; pre[tail] = head; choose[tail] = i; if (num == final) break; } } if (vis[final]) { int len = 0; printf("%d/n", best[tail]); while (tail != 0) { len++; path[len] = choose[tail]; tail = pre[tail]; } for (int i = len; i > 0; i--) { printf("%c", path[i]+'A'); len++; if (len % 60 == 0 || i == 1) printf("/n"); } } head++; } return 0; }

 

題目六：Sweet Butter

算法：SPFA

題目大意：給出一張圖，有一些人在某些點，求出最小的集合代價。

由於這張圖邊比較稀疏，用鄰接表存好，且訪問起來更方便；

用SPFA分別求出任意兩點間的最短路。

/* ID: zhangji42 TASK: butter LANG: C++ */ #include <stdio.h> #include <string.h> const int maxn = 801, maxcow = 501, maxedge = 1451*2; int cow[maxcow], first[maxn]; int v[maxedge], w[maxedge], next[maxedge]; int q[maxedge*5], best[maxn], vis[maxn]; int main(){ freopen("butter.in", "r", stdin); freopen("butter.out", "w", stdout); int M, N, C; scanf("%d %d %d", &M, &N, &C); for (int i = 1; i <= M; i++) scanf("%d", &cow[i]); memset(first,-1,sizeof(first)); int u; for (int i = 1; i <= C; i++) { scanf("%d %d %d", &u, &v[i], &w[i]); next[i] = first[u]; first[u] = i; v[C+i] = u; w[C+i] = w[i]; next[C+i] = first[v[i]]; first[v[i]] = C+i; } int min = 1 << 30; for (int i = 1; i <= N; i++) { int head = 0, tail = 0, now; q[head] = i; memset(vis,0,sizeof(vis)); vis[i] = 1; for (int j = 1; j <= N; j++) best[j] = 1 << 29; best[i] = 0; while (head <= tail) { now = q[head]; for (int k = first[now]; k != -1; k = next[k]) if (best[now] + w[k] < best[v[k]]) { best[v[k]] = best[now] + w[k]; if (!vis[v[k]]) { q[++tail] = v[k]; vis[v[k]] = 1; } } vis[now] = 0; head++; } now = 0; for (int j = 1; j <= M; j++) {now += best[cow[j]];}//printf("%d %d/n", cow[j], best[cow[j]]);} if (now < min) min = now; } printf("%d/n", min); return 0; }

 

很好，這一個星期的任務完成了，不過接下來還要奮戰學校的OJ啊！加油！