#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
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
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;
}
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时间： 2024-08-04 21:57:43