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re: NOIP2012 �U�中行记

zoushihuan — Sat, 24 Nov 2012 12:27:00 GMT

��g��是beizhong的，老huang的student�Q?
I'm chusan。who are you�Q?img src ="http://m.shnenglu.com/Climber-pI/aggbug/195636.html" width = "1" height = "1" />

zoushihuan 2012-11-24 20:27 发表评论

re: NOIp 2005 �q�河

PZYR — Thu, 08 Sep 2011 05:30:00 GMT

#include
#include
using namespace std;
#define MAXN 1000000;
bool L[200000] = {0};
int f[200000] = {0}, stone[110] = {0};
int min(int x, int y){return x < y ? x : y;}
void swap(int x, int y){
int k = stone[x];
stone[x] = stone[y];
stone[y] = k;
}
int main(){
int l, S, T, M, i, j;
scanf("%d%d%d%d", &l, &S, &T, &M);
for (i = 1; i <= M; i++) scanf("%d", &stone[i]);
stone[0] = 0;
for (i = 1; i < M; i++)
for (j = i+1; j <= M; j++)
if (stone[i] > stone[j]) swap(i, j);
stone[++M] = l;
for (i = 1; i <= M; i++){
while (stone[i] - stone[i-1] > 2520) stone[i] -= 2520;
if (i != M) L[stone[i]] = 1;
}
l = stone[M--];
for (i = 0; i <= l; i++) f[i] = MAXN; f[0] = 0;
for (i = 0; i < l; i++)
for (j = S; j <= T; j++){
int k = i+j;
if (i + j >= l) k = l;
f[k] = min(f[k], f[i]+L[i]);
}
printf("%d\n", f[l]);
}

PZYR 2011-09-08 13:30 发表评论

re: Dp札记[未登录]

Climber.pI — Wed, 18 May 2011 11:14:00 GMT

我发现我只有上面�c�d��的dp我会�?..看而不写基本全�?img src ="http://m.shnenglu.com/Climber-pI/aggbug/146688.html" width = "1" height = "1" />

Climber.pI 2011-05-18 19:14 发表评论

re: Prepare NOIp 2010 Final Plan

Aha — Fri, 19 Nov 2010 10:04:00 GMT

额~发现在现在书便宜了。。记得我当时买的什么来着�Q�两本就70++了，唔唔�?img src ="http://m.shnenglu.com/Climber-pI/aggbug/134101.html" width = "1" height = "1" />

Aha 2010-11-19 18:04 发表评论

re: Prepare NOIp 2010 Final Plan

Climber.pI — Mon, 25 Oct 2010 13:55:00 GMT

@dementrock
拿来热��n - -

Climber.pI 2010-10-25 21:55 发表评论

re: Prepare NOIp 2010 Final Plan

dementrock — Sun, 24 Oct 2010 09:13:00 GMT

usaco铜组��免�?太水没啥价�?img src ="http://m.shnenglu.com/Climber-pI/aggbug/131066.html" width = "1" height = "1" />

dementrock 2010-10-24 17:13 发表评论

re: NOIp 2005 �q�河

ftiasch — Sat, 09 Oct 2010 00:53:00 GMT

2520恰好�?, 2, ..., 10的最��公倍数。原理就是可以证明说状态函数的��D��定会出现大段的重复。在理论上可以保证的��是2520。更多的分析应该用裴蜀定理来做�?img src ="http://m.shnenglu.com/Climber-pI/aggbug/129156.html" width = "1" height = "1" />

ftiasch 2010-10-09 08:53 发表评论

Climber.pI — Fri, 08 Oct 2010 13:57:00 GMT

@dementrock
不会爆栈�?

Climber.pI 2010-10-08 21:57 发表评论

dementrock — Wed, 06 Oct 2010 02:31:00 GMT

记忆化搜索是不可能有未计��的……只可能写猥�?img src ="http://m.shnenglu.com/Climber-pI/aggbug/128763.html" width = "1" height = "1" />

dementrock 2010-10-06 10:31 发表评论