uva 10254 - The Priest Mathematician(dp)

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题目链接：uva 10254 - The Priest Mathematician

题目大意：四根柱子的汉诺塔问题，问说移动的最少次数。

解题思路：一开始想着说用两个大数数组f[N], b[N], 然后一定存在一个值k，使得f[i] = f[k] * 2 + b[i - k]最小，然后b[i] = b[i - 1] * 2+ 1,但是这样枚举k的计算量非常大。

然后参考别人打表找出来的规律，每次增长的个数为2个2^1,3个2^2,4个2^3.....m - 1个2^m。

#include <stdio.h>#include <string.h>#define max(a,b) (a)>(b)?(a):(b)const int N = 10005;const int M = 400;int n, f[N][M], d[M];void put(int c[]) {for (int i = c[0]; i; i--)printf("%d", c[i]);printf("\n");}void mul(int c[]) {int sum = 0;for (int i = 1; i <= c[0]; i++) {sum += c[i] * 2;c[i] = sum %  10;sum /= 10;}while (sum) {c[++c[0]] = sum % 10;sum /= 10;}}void add(int ans[], int a[], int b[]) {int sum = 0;ans[0] = max(a[0],  b[0]);for (int i = 1; i <= ans[0]; i++) {if (i <= a[0]) sum += a[i];if (i <= b[0]) sum += b[i];ans[i] = sum % 10;sum /= 10;}while (sum) {ans[++ans[0]] = sum % 10;sum /= 10;}}void init() {f[1][0] = f[1][1] = d[0] = 1;d[1] = 2;int c = 0, top = 2;for (int i = 2; i <= 10000; i++) {add(f[i], f[i - 1], d);c++;if (c == top) {top++, c = 0;mul(d);}}}int main () {init();while (scanf("%d", &n)  == 1) {if (n)put(f[n]);else printf("0\n");}return 0;}