SGU 403

#include<stdio.h>
int x;
int main(){
while(~scanf("%d",&x))
printf("%d\n", 2*x+1);
return 0;
}

#include<iostream>
#include<cstdio>
#include<vector>
#include<string.h>
using namespace std;
int n, m;
char s[105][105];
int main(){
    int i ,j;
    while(~scanf("%d %d",&m,&n))
    {
        m--;
        for(i = 0; i < n; i++)scanf("%s", s[i]);
        printf("%s\n", s[m%n]);
    }

    return 0;
}

#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 105;
int s[N];
int main() {
	int n, m, a, b, x, y;
	scanf("%d%d", &n, &m);
	for(int i = 0; i < m; i ++) {
		scanf("%d%d", &a, &b);
		for(int j = 0; j < n; j ++) {
			scanf("%d%d", &x ,&y);
			if(a == x) s[j]++;
			if(y == b) s[j]++;
			if(a-b == x-y) s[j] += 3;
			
			if((a-b)*(x-y) > 0) s[j] += 2;
			if(a == b && x == y) s[j] += 2;
		}
	}
	for(int i = 0; i < n; i ++) {
		printf("%d%c", s[i], i == n-1 ? '\n' : ' ');
	}
	return 0;
}

#include<iostream>
#include<cstdio>
#include<vector>
#include<string.h>
using namespace std;
int n, m;
int a[12][101];
vector<int>A[11], tmp, ans, no;
int main(){
    int i ,j, u, v;
    while(~scanf("%d %d",&n,&m))
    {
        for(i = 1; i <= n; i++)
        {
            memset(a[i], 0, sizeof a[i]);
            A[i].clear();
            scanf("%d",&j);
            while(j--)
            {
                scanf("%d",&u);
                A[i].push_back(u);
                a[i][u]++;
            }
        }
        while(m--)
        {
            scanf("%d", &u);
            tmp.clear();
            ans.clear();
            no.clear();
            memset(a[0], 0, sizeof a[0]);
            while(u--)
            {
                scanf("%d",&v);
                if(v<0){no.push_back(-v); continue;}
                tmp.push_back(v);
            }
            for(i = 1; i <= n; i++)
            {
                bool ok = true;
                for(j = 0; ok && j < no.size(); j++)
                if(a[i][no[j]])ok = false;

                for(j = 0; ok && j < tmp.size(); j++)
                if(!a[i][tmp[j]])ok = false;

                if(ok)ans.push_back(i);
            }
            printf("%d\n", ans.size());
            for(i = 0; i < ans.size(); i++)
            {
                u = ans[i];
                printf("%d ", A[u].size());
                for(j = 0; j < A[u].size(); j++)
                    printf("%d%c", A[u][j], (j+1)==A[u].size()?'\n':' ');
            }
        }
    }

    return 0;
}

Java大数 

题解

SGU 408

显然是2个数字越接近越好，即平方数。。而且答案一定为整数

#include<iostream>
#include<cstdio>
#include<vector>
#include<string.h>
using namespace std;
#define ll long long
#define N 10000
ll n;
int main(){
    while(cin>>n){
        if(n<=1){ cout<<n; puts(".000"); continue;}
        ll ans = 5, x = 2, y = 2;
        bool yes = true;
        for(ll i = 3; i <= n; i++)
        {
            if(yes) x++;
            else y++;
            yes ^= 1;
            ans += x*y;
        }
        cout<<ans; puts(".000");
    }
    return 0;
}

#include <cstdio>
#include <cstring>

using namespace std;
const int N = 25;
char a[N][N];
char ans[N*N][N*N];
int main(){
<span style="white-space:pre">	</span>int n, k;
    scanf("%d%d", &n, &k);
    for(int i = 0; i < n; i ++) {
    	for(int j = 0; j < n; j ++) {
    		if(k > 0) {
    			a[i][j] = '*';
    			k--;
    		} else a[i][j] = '.';
    	}
    }
    
    for(int i = 0; i < n; i ++) {
    	for(int j = 0; j < n; j ++) {
    		for(int ii = 0; ii < n; ii ++) {
    			for(int jj = 0; jj < n; jj ++) {
    				ans[i*n+ii][j*n+jj] = a[(ii + j) % n][(jj + i) % n];
    			}
    		}
    	}
    }
    
    for(int i = 0; i < n*n; i ++) {
    	for(int j = 0; j < n*n; j ++) {
    		printf("%c", ans[i][j]);
    	} puts("");
    }
    return 0;
}

百度有解。。

减到使得最大数是最小数的2倍就停止，然后翻倍再减

SGU 411

后缀数组

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX_N = 90007;

#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)

int wa[MAX_N], wb[MAX_N], wv[MAX_N], ws[MAX_N];
int height[MAX_N], rank[MAX_N], sa[MAX_N];

struct Suffix {
	int c0(int *r, int a, int b) {
		return r[a] == r[b] && r[a+1] == r[b+1] && r[a+2] == r[b+2];
	}
	int c12(int k, int *r, int a, int b) {
		if(k==2)
		  return r[a] < r[b] || r[a] == r[b] && c12(1, r, a+1, b+1);
		else 
		  return r[a] < r[b] || r[a] == r[b] && wv[a+1] < wv[b+1];
	}
	void sort(int *r,int *a,int *b,int n,int m)	{
		int i;
		for(i = 0; i < n; i++) wv[i] = r[a[i]];
		for(i = 0; i < m; i++) ws[i] = 0;
		for(i = 0; i < n; i++) ws[wv[i]]++;
		for(i = 1; i < m; i++) ws[i] += ws[i-1];
		for(i = n-1; i >= 0; i--) b[--ws[wv[i]]] = a[i];

	}
	void dc3(int *r, int *sa, int n, int m) {
		int i, j, *rn = r + n, *san = sa + n, ta = 0, tb = (n+1) / 3,tbc = 0, p;
		r[n] = r[n+1] = 0;
		for(i = 0; i < n; i++) if(i % 3 != 0) wa[tbc++] = i;
		sort(r + 2, wa, wb, tbc, m);
		sort(r + 1, wb, wa, tbc, m);
		sort(r, wa, wb, tbc, m);
		for(p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++)
			rn[F(wb[i])] = c0(r, wb[i-1], wb[i]) ? p - 1 : p++;
		if(p < tbc) dc3(rn, san, tbc, p);
		else for(i = 0; i < tbc; i++) san[rn[i]]=i;	
		for(i = 0; i < tbc; i++) if(san[i] < tb) wb[ta++] = san[i]*3;
		if(n % 3 == 1) wb[ta++] = n-1;
		sort(r, wb, wa, ta, m);
		for(i = 0; i < tbc; i++) wv[wb[i] = G(san[i])] = i;
		for(i = 0, j = 0, p = 0; i < ta && j < tbc; p++)
		sa[p] = c12(wb[j]%3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];
		for(; i<ta; p++) sa[p] = wa[i++];
		for(; j<tbc; p++) sa[p] = wb[j++];
	}
	void calheight(int *r, int *sa, int n) {
		for (int i = 1; i <= n; ++i) rank[sa[i]] = i;
		for (int i = 0, k = 0; i < n; ++i) {
			if(k) --k;
			int j = sa[rank[i]-1];
			while (i + k < n && j + k < n && r[i+k] == r[j+k]) ++k;
			height[rank[i]] = k; 
		}
	}
};

int maxLen, begin;
char str[MAX_N], s1[MAX_N], s[MAX_N];
int d[MAX_N], p[MAX_N];
char m[MAX_N];

void Manacher(int st, int len) {
	int cnt = 0;
	m[cnt++] = '$', m[cnt++] = '#';
	int len1 = min(st + len, (int)strlen(s1));
	for (int i = st; i < len1; ++i) {
		m[cnt++] = s1[i];
		m[cnt++] = '#';
	}
	m[cnt] = '\0';
	memset(p, 0, sizeof p);
	int mx = 0, id = 0;
	for (int i = 1; i < cnt; ++i) {
		if (mx > i) p[i] = min(p[2 * id - i], mx - i);
		else p[i] = 1;
		while (i - p[i] > 0 && i + p[i] < cnt && m[i - p[i]] == m[i + p[i]]) ++p[i];
		if (i + p[i] > mx) {
			mx = i + p[i];
			id = i;
		}
	}
	int d = 0;
	for (int i = 1; i < cnt; ++i) {	
		if (p[i] - 1 > maxLen) {
			maxLen = p[i] - 1;
			begin = st - (p[i] - 1) / 2 + d;
		}			
		if (m[i] != '#') ++d;
	}
}

int main() {
	scanf("%s", s1);
	int l1 = (int) strlen(s1);
	int cnt = 0;
	for (int i = 0; s1[i]; ++i) {
		str[cnt++] = s1[i];
	}
	str[cnt++] = 'z' + 1;
	scanf("%s", s);
	int l2 = (int) strlen(s);
	for (int i = 0; s[i]; ++i) {
		str[cnt++] = s[i];
	}
	str[cnt] = '\0';
	int n = l1 + l2 + 1;
	for (int i = 0; i < n; ++i) {
		d[i] = str[i] - 'a' + 1;
	}
	d[n] = 0;
	Suffix ans;
	ans.dc3(d, sa, n + 1, 30);
	ans.calheight(d, sa, n);
	maxLen = 0, begin = 0;
	for (int i = 1; i <= n; ++i) {
		if (sa[i - 1] < l1 && sa[i] > l1) {
			Manacher(sa[i - 1], height[i]);
		}
		if (sa[i - 1] > l1 && sa[i] < l1) {
			Manacher(sa[i], height[i]);
		}
	}
	for (int i = 0; i < maxLen; ++i)
	  putchar(s1[i + begin]);
	puts("");
	return 0;
}

枚举起点，然后任意2个点先成对，剩下的孤立点再投入相邻的集合里

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int MAX_N = 107;

int n, m;
bool vis[MAX_N];
vector<int> edge[MAX_N], cou[MAX_N];
int bel[MAX_N], sig;

void addedge(int u, int v) {
	edge[u].push_back(v);
	edge[v].push_back(u);
}

void clear() {
	sig = 0;
	memset(vis, false, sizeof vis);
	memset(bel, 0, sizeof bel);
	for (int i = 0; i <= n; ++i)
	  cou[i].clear();
}

void dfs(int u) {
	vis[u] = true;
	cou[u].push_back(u);
	for (int i = 0; i < (int) edge[u].size(); ++i) {
		if (!vis[edge[u][i]]) {
			dfs(edge[u][i]);
			if (cou[edge[u][i]].size() == 1) {
				cou[u].push_back(edge[u][i]);
			}
		}
	}
	if (cou[u].size() > 1) {
		++sig;
		for (int i = 0; i < (int)cou[u].size(); ++i) {
			bel[cou[u][i]] = sig;
		}
	}
}

bool check() {
	for (int i = 0; i < n; ++i) {
		if (bel[i] == 0) return false;
	}
	return true;
}

int main() {
	int T;
	scanf("%d", &T);
	while (T-- > 0) {
		scanf("%d%d", &n, &m);

		for (int i = 0; i < 103; ++i) 
		  edge[i].clear();
		memset(bel, 0, sizeof bel);

		for (int i = 0; i < m; ++i) {
			int u, v;
			scanf("%d%d", &u, &v);
			--u, --v;
			addedge(u, v);
		}
		for (int i = 0; i < n; ++i) {
			clear();
			dfs(i);
			if (check()) {
				for (int j = 0; j < n; ++j) {
					if (j == 0) printf("%d", bel[j]);
					else printf(" %d", bel[j]);
				}
				puts("");
				break;
			}
		}
	}
	return 0;
}