USACO 2.2

前言

然而这次已经变成了动态规划

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洛谷 1465 序言页码 Preface Numbering

题目

求$[1\dots n]$的罗马数字中各个罗马数字的个数

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分析

然而竟然发现可以拆成十进制的每一位
具体是这样的
$I,II,III,IV,V,VI,VII,VIII,IX$(最后一位)
$X,XX,XXX,XL,L,LX,LXX,LXXX,XC$(倒数第二位)
$C,CC,CCC,CD,D,DC,DCC,DCCC,CM$(倒数第三位)
$M,MM,MMM$(首位)
然后暴力代码就出来了

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代码

/*
ID:lemondi1
LANG:C++
TASK:preface
*/
#include <cstdio>
#define rr register
using namespace std;
const char b[7]={'I','V','X','L','C','D','M'};
int n,a[7];
void add(int now,int x){
    a[now-1<<1]+=(x==1)+(x==2)*2+(x==3)*3+(x==4)+(x==6)+(x==7)*2+(x==8)*3+(x==9);
	if (now<4) a[now-1<<1|1]+=(x==4)+(x==5)+(x==6)+(x==7)+(x==8),a[now<<1]+=(x==9);
}
void frec(int x){rr int now=0; while (x) add(++now,x%10),x/=10;}
signed main(){
	freopen("preface.in","r",stdin);
	freopen("preface.out","w",stdout);
	scanf("%d",&n);
	for (rr int i=1;i<=n;++i) frec(i);
	for (rr int i=0;i<7;++i) if (a[i]) printf("%c %d\n",b[i],a[i]);
    return 0;
}

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洛谷 1466 集合 Subset Sums

题目

把$[1\dots n]$分成2份，使和分别相等，问有多少种方案

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分析

其实这道题就是01背包，$f[j]=f[j-i](1\le i\le n,m\ge j\ge i)$

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代码

/*
ID:lemondi1
LANG:C++
TASK:subset
*/
#include <cstdio>
#define rr register
using namespace std;
unsigned n,dp[751];
signed main(){
	freopen("subset.in","r",stdin);
	freopen("subset.out","w",stdout);
	scanf("%u",&n); dp[0]=1;
	if ((n*n+n>>1)&1) return !printf("0\n");//根本不可能分开
	for (rr unsigned i=1;i<=n;++i)
	for (rr unsigned j=n*n+n>>2;j>=i;--j) dp[j]+=dp[j-i];
    return !printf("%u\n",dp[n*n+n>>2]>>1);//记得除以2
}

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洛谷 1467 循环数 Runaround Numbers

分析

老老实实按照题目打就行了

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代码

/*
ID:lemondi1
LANG:C++
TASK:runround
*/
#include <cstdio>
#include <cstring>
#define rr register
using namespace std;
unsigned n,b[16]; 
signed allright(unsigned n){
	rr unsigned len=0,t=n,v=0;
	while (t) b[++len]=t%10,t/=10; t=1;
	for (rr int i=1;i<=len;++i){
		if (v&(1<<b[len-t+1]-1)||!b[len-t+1]) return 0;
		v|=(1<<b[len-t+1]-1);
		t=(t+b[len-t+1]-1)%len+1;
	}
	return t==1;
}
signed main(){
    freopen("runround.in","r",stdin);
    freopen("runround.out","w",stdout);
    scanf("%u",&n);
    while (++n)
        if (allright(n)) 
		    return !printf("%d\n",n);
}

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洛谷 1468 派对灯 Party Lamps

分析

然而可以发现组合方式只有1,2,3,4,1+4,2+4,3+4,nothing
也可以发现一定是长度为6的循环节，所以说代码就出来了

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代码

/*
ID:lemondi1
LANG:C++
TASK:lamps
*/
#include <cstdio>
#define rr register
using namespace std;
const int a[8]={0,14,21,27,36,42,49,63};
int b[6]={-1,-1,-1,-1,-1,-1},n,c,flag;
void ok(int c){
	for (rr int i=0;i<6;++i) if (b[i]!=-1&&((a[c]>>5-i)&1)!=b[i]) return; flag=1;
	for (rr int i=0;i<n;++i) putchar(48+((a[c]>>5-(i%6))&1)); putchar(10);
}
signed main(){
	freopen("lamps.in","r",stdin);
	freopen("lamps.out","w",stdout);
	scanf("%d%d",&n,&c);
	while (1){
		rr int t;
		scanf("%d",&t);
		if (t==-1) break;
		b[(t-1)%6]=1;
	}
	while (1){
		rr int t;
		scanf("%d",&t);
		if (t==-1) break;
		b[(t-1)%6]=0;
	}
	if (!c) {ok(7); if (flag) return 0;}
	else if (c==1) {ok(0),ok(2),ok(3),ok(5); if (flag) return 0;}
	else if (c==2) {ok(0),ok(1),ok(2),ok(4),ok(5),ok(6),ok(7); if (flag) return 0;}
	else {ok(0),ok(1),ok(2),ok(3),ok(4),ok(5),ok(6),ok(7); if (flag) return 0;}
	printf("IMPOSSIBLE\n");
	return 0;
}