nim遊戲的必勝策略（博弈論）

转载自：http://www.cnblogs.com/vongang/archive/2013/03/10/2952978.html POJ【数论/组合/博弈论】题目列表

果然dp還是弱項啊啊啊啊．． 不過比最開始的完全無從下手強了不少應該．．． 至少dp狀態表示相對了．．．．轉移方程沒想出來嗚嗚嗚 官方題解：设d(i, j)d(i,j)表示前ii个数，模pp为jj的方案数，则容易得到d(0, 0)=1, d(i, j)=d(i-1, j)+sum_{j=0}^{p-1} d(i-1, (j-a[i]) mod p)d(0,0)=1,d(i,j)=d(i−1,j)+∑j=0p−1d(i−1,(j−a[i]) mod p)，很多人没1a是因为没注意|a_i| le 10^9∣ai∣≤109

贪心．．尽量把一样的材料放在一起．．． 然后写蠢了．．妈蛋．．． 详情见代码 代码实现 1/************************************************************************* 2 > File Name: code/bc/#56/1001.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年09月19日 星期六 18时55分49秒 6 ************************************************************************/ 7 8#include<iostream> 9#include<iomanip> 10#include<cstdio> 11#include<algorithm> 12#include<cmath> 13#include<cstring> 14#include<string> 15#include<map> 16#include<set> 17#include<queue> 18#include<vector> 19#include<stack> 20#include<cctype> 21#define y1 hust111qqz 22#define yn hez111qqz 23#define j1 cute111qqz 24#define ms(a,x) memset(a,x,sizeof(a)) 25#define lr dying111qqz 26using namespace std; 27#define For(i, n) for (int i=0;i<int(n);++i) 28typedef long long LL; 29typedef double DB; 30const int inf = 0x3f3f3f3f; 31const int N=5E2+7; 32int a[N],b[N]; 33int ans,cnt; 34int p[N]; 35int n; 36int main() 37{ 38 #ifndef ONLINE_JUDGE 39 freopen("in.txt","r",stdin); 40 #endif 41 int T; 42 cin>>T; 43 while (T--) 44 { 45 ans = 0; 46 cnt = 0; 47 ms(p,0); 48 scanf("%d",&n); 49 for ( int i = 0 ; i < n ; i++ ) 50 { 51 scanf("%d %d",&a[i],&b[i]); 52 p[a[i]]+=b[i]; 53 } 54 int kind = 0; 55 for ( int i = 1 ;i <= 500 ; i++) 56 { 57 if (p[i]!=0) 58 { 59 // cnt = cnt + (p[i]-1)/64 + 1; 60 cnt = cnt + (p[i]+63)/64;　//这样写不知高到哪里去了． 61 p[i] = 0; 62 } 63 64 } 65 //ans = ans + (cnt-1)/36+1; 66 ans = (cnt+35)/36; //不知高到哪里去了．．． 67 printf("%d\n",ans); 68 69 70 } 71 72 73 #ifndef ONLINE_JUDGE 74 fclose(stdin); 75 #endif 76 return 0; 77}

x的二进制表示中1的个数即为答案． 原因是，每天晚上糖果数量翻倍，相当于左移１位，这时候二进制表示中1的数量不变 也就是说，二进制表示中的所有的1，一定都是添加进去的 而且也只有二进制表示中的1是添加进去的

初识分快. 引一段题解： Let’s split rectangle 106 × 106 by vertical lines into 1000 rectangles 103 × 106. Let’s number them from left to right. We’re going to pass through points rectangle by rectangle. Inside the rectangle we’re going to pass the points in increasing order of y-coordinate if the number of rectangle is even and in decreasing if it’s odd. Let’s calculate the maximum length of such a way. The coordinates are independent. By y-coordinate we’re passing 1000 rectangles from0 to 106, 109 in total. By x-coordinate we’re spending 1000 to get to the next point of current rectangle and 2000 to get to next rectangle. That means, 2 * 109 + 2000000 in total, which perfectly fits.