USACO 2.2

Posted on 2011-11-28 18:19 MDeath-Kid 阅读(154) 评论(0) 收藏举报

PROB Preface Numbering [ANALYSIS] ---- 又是枚举，让我不得不相信枚举的实力了。

应该是估算数据量的作用，1000的数量级，不过才怪，我还想着去按位DP。。。。

int n;
int sum_ge[7];
//	idx 1 - 3
void calculate(int x,int idx)
{
	switch(x) {
		case 0:							break;
		case 1:	sum_ge[2*(idx-1)]++;				break;
		case 2:	sum_ge[2*(idx-1)]+=2;				break;
		case 3:	sum_ge[2*(idx-1)]+=3;				break;
		case 4:	sum_ge[2*(idx-1)]+=1;sum_ge[2*(idx-1)+1]++;	break;
		case 5:	sum_ge[2*(idx-1)+1]++;				break;
		case 6:	sum_ge[2*(idx-1)]+=1;sum_ge[2*(idx-1)+1]++;	break;
		case 7:	sum_ge[2*(idx-1)]+=2;sum_ge[2*(idx-1)+1]++;	break;
		case 8:	sum_ge[2*(idx-1)]+=3;sum_ge[2*(idx-1)+1]++;	break;
		case 9:	sum_ge[2*(idx-1)]+=1;sum_ge[2*(idx-1)+2]++;	break;
	}
}

int main()
{
	FOPENTI
	FOPENTO
	SET(sum_ge,0);
	SCF(n);
	FOR(i,1,n) {
		int t=1,x=i;
		while(x) {
			calculate(x%10,t++);
			x /= 10;
		}
	}
	if(sum_ge[0]) printf("I %d\n",sum_ge[0]);
	if(sum_ge[1]) printf("V %d\n",sum_ge[1]);
	if(sum_ge[2]) printf("X %d\n",sum_ge[2]);
	if(sum_ge[3]) printf("L %d\n",sum_ge[3]);
	if(sum_ge[4]) printf("C %d\n",sum_ge[4]);
	if(sum_ge[5]) printf("D %d\n",sum_ge[5]);
	if(sum_ge[6]) printf("M %d\n",sum_ge[6]);
}

USER: Rain M [m3324631]
TASK: preface
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 3048 KB]
   Test 2: TEST OK [0.000 secs, 3048 KB]
   Test 3: TEST OK [0.000 secs, 3048 KB]
   Test 4: TEST OK [0.000 secs, 3048 KB]
   Test 5: TEST OK [0.000 secs, 3048 KB]
   Test 6: TEST OK [0.000 secs, 3048 KB]
   Test 7: TEST OK [0.000 secs, 3048 KB]
   Test 8: TEST OK [0.000 secs, 3048 KB]

All tests OK.
YOUR PROGRAM ('preface') WORKED FIRST TIME!  That's fantastic
-- and a rare thing.  Please accept these special automated
congratulations.

Here are the test data inputs:

------- test 1 ----
1
------- test 2 ----
20
------- test 3 ----
100
------- test 4 ----
500
------- test 5 ----
1000
------- test 6 ----
2974
------- test 7 ----
3213
------- test 8 ----
3499
Keep up the good work!
Thanks for your submission!

PROB Subset Sums [ANALYSIS] ---- 明显就是背包，浆糊了。。

// solution 1		求和，DP
//	solution 1
//	1：	求和，看是否是整数
//	2：	是的话 DP 和的一半，有多少种组成

//	不可取重复，纠结了很长时间，终于明白了
//	这就是一般背包
int n,sumhalf,ans;
LLong anst[MAXN];



int main()
{
	FOPENTI
	FOPENTO
        SCF(n);
        int sum = ((n+1)*n)/2;
        if(sum & 0x1) puts("0");
        else {
        	anst[0] = 1;
		for(int i = 1;i<=n;i++) {
			for(int j = sum/2;j>=i;j--) {
				anst[j] += anst[j-i];
			}
		}
		printf("%lld\n",anst[sum/2]/2);
        }

}

USER: Rain M [m3324631]
TASK: subset
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 3052 KB]
   Test 2: TEST OK [0.000 secs, 3052 KB]
   Test 3: TEST OK [0.000 secs, 3052 KB]
   Test 4: TEST OK [0.000 secs, 3052 KB]
   Test 5: TEST OK [0.000 secs, 3052 KB]
   Test 6: TEST OK [0.000 secs, 3052 KB]
   Test 7: TEST OK [0.000 secs, 3052 KB]

All tests OK.
YOUR PROGRAM ('subset') WORKED FIRST TIME!  That's fantastic
-- and a rare thing.  Please accept these special automated
congratulations.

Here are the test data inputs:

------- test 1 ----
7
------- test 2 ----
15
------- test 3 ----
24
------- test 4 ----
31
------- test 5 ----
36
------- test 6 ----
39
------- test 7 ----
37
Keep up the good work!
Thanks for your submission!

PROB Runaround Numbers [ANALYSIS] ---- 枚举，题目看了好几遍。

限制条件挺多，每位数字不能重复，不能有0，这个数只能是一个环！

unsigned int n;
bool flag;
bool Ans[MAXN];

bool check(unsigned int x)
{
	char ss[MAXN];
	SET(Ans,false);
	int chee = 0;
	sprintf(ss,"%u",x);
	int len = strlen(ss);
	F(i,len) {
		if(chee & (1<<(ss[i]-'0'))) return false;
		chee |= (1<<(ss[i]-'0'));
	}
	for(int i = 0;;) {
		int k = ss[i] - '0',che = i;
		if(k == 0) return false;
		if(Ans[i]) {
			if(i == 0) flag = 1;
			else return false;

			break;
		}
		else Ans[i] = true;
		while(k--) {
			if(++i >= len) i = 0;
		}
		if(ss[i] == ss[che]) return false;
	}
	F(i,len) {
		if(!Ans[i]) return false;
	}
	if(flag)
	return true;
}


int main()
{
        FOPENTI
        FOPENTO
	scanf("%u",&n);
	while(!check(++n));
	printf("%u\n",n);
}

USER: Rain M [m3324631]
TASK: runround
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 3048 KB]
   Test 2: TEST OK [0.011 secs, 3048 KB]
   Test 3: TEST OK [0.000 secs, 3048 KB]
   Test 4: TEST OK [0.011 secs, 3048 KB]
   Test 5: TEST OK [0.108 secs, 3048 KB]
   Test 6: TEST OK [0.065 secs, 3048 KB]
   Test 7: TEST OK [0.162 secs, 3048 KB]

All tests OK.
YOUR PROGRAM ('runround') WORKED FIRST TIME!  That's fantastic
-- and a rare thing.  Please accept these special automated
congratulations.

Here are the test data inputs:

------- test 1 ----
99
------- test 2 ----
111110
------- test 3 ----
134259
------- test 4 ----
348761
------- test 5 ----
1000000
------- test 6 ----
5000000
------- test 7 ----
9000000
Keep up the good work!
Thanks for your submission!

PROB Party Lamps [ANALYSIS] ---- 模拟

struct str_hash {
        size_t operator()(const string& str) const        {
                return __stl_hash_string(str.c_str());
	}
};

int n,stepgoal;
char goal[105],status[105];
map<string,int> has;
vector<string> vec;


void set_bit(int x,int bol)
{
        goal[x-1] = bol + '0';
}

bool findans()
{
        F(i,n) {
                if(goal[i] == '3') continue;
                if(status[i] != goal[i]) return 0;
        }
        return 1;
}


void dfs(int depth)
{
        if(findans()) {
        	vec.push_back(status);
        }
        if(depth == stepgoal) {
        	return;
        }
        //1
        F(i,n) status[i] ='1'-status[i]+'0';
        if(!has[status])
        has[status] = 1,dfs(depth + 1);
        F(i,n) status[i] ='1'-status[i]+'0';

        //2
        for(int i = 0;i<n;i+=2) {
        	status[i] ='1'-status[i]+'0';
        }
        if(!has[status])
        has[status] = 1,dfs(depth + 1);
        for(int i = 0;i<n;i+=2) {
        	status[i] ='1'-status[i]+'0';
        }
        //3
        for(int i = 1;i<n;i+=2) {
        	status[i] ='1'-status[i]+'0';
        }
        if(!has[status])
        has[status] = 1,dfs(depth + 1);
        for(int i = 1;i<n;i+=2) {
        	status[i] ='1'-status[i]+'0';
        }
        //4
        for(int i = 0;i<n;i+=3) {
        	status[i] ='1'-status[i]+'0';
        }
        if(!has[status])
        has[status] = 1,dfs(depth + 1);
        for(int i = 0;i<n;i+=3) {
        	status[i] ='1'-status[i]+'0';
        }
}


int main()
{
	FOPENTI
	FOPENTO
        SCFD(n,stepgoal);
        SET(goal,'\0');
        F(i,n) goal[i]='3';
        int a;
        SCF(a);
        while(a != -1) {
                set_bit(a,1);
                SCF(a);
        }
        SCF(a);
        while(a != -1) {
                set_bit(a,0);
                SCF(a);
        }
        SET(status,'\0');
        F(i,n)status[i] = '1';
        has.clear();
        has[status] = 1;
        dfs(0);

        sort(vec.begin(),vec.end());

        F(i,vec.size()) {
                cout<<vec[i]<<endl;
        }
        if(vec.size() == 0) {
		puts("IMPOSSIBLE");
        }
}

一开始忘记没有答案的时候了。WA了一次

USER: Rain M [m3324631]
TASK: lamps
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 3196 KB]
   Test 2: TEST OK [0.000 secs, 3196 KB]
   Test 3: TEST OK [0.000 secs, 3196 KB]
   Test 4: TEST OK [0.000 secs, 3196 KB]
   Test 5: TEST OK [0.000 secs, 3196 KB]
   Test 6: TEST OK [0.000 secs, 3196 KB]
   Test 7: TEST OK [0.000 secs, 3196 KB]
   Test 8: TEST OK [0.000 secs, 3196 KB]

All tests OK.
Your program ('lamps') produced all correct answers!  This is your
submission #2 for this problem.  Congratulations!

Here are the test data inputs:

------- test 1 ----
10
0
-1
-1
------- test 2 ----
10
0
-1
1 -1
------- test 3 ----
20
3
-1
1 3 5 -1
------- test 4 ----
50
100
1 -1
-1
------- test 5 ----
75
250
-1
-1
------- test 6 ----
100
8394
1 7 13 19 25 31 37 43 49 55 -1
64 -1
------- test 7 ----
100
2000
31 86 23 -1
42 -1
------- test 8 ----
100
8950
-1
-1
Keep up the good work!
Thanks for your submission!

good luck

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USACO 2.2