差别

这里会显示出您选择的修订版和当前版本之间的差别。

--- 2020-2021:teams:die_java:front_page_summertrain1 [2020/07/17 17:43]
fyhssgss [A.B-Suffix Array]
+++ 2020-2021:teams:die_java:front_page_summertrain1 [2020/10/17 22:58] (当前版本)
fyhssgss
@@ 行 254: / 行 254: @@
 故变成： $$
-约束条件：x^TAx\leq1
+约束条件：x^TAx=1
 \\ 目标函数：\{b^T·x\}
 $$ 考虑拉格朗日数乘法 $$
@@ 行 343: / 行 343: @@
     }
     return 0;
+}
+</code>
+</hidden>
+===== F.Infinite String Comparision =====
+=== 题意 ===
+将两个串无限循环，问是否相等
+=== 题解 ===
+solved by hxm
+容易发现，只需要判断最长串的两个周期内是否相等即可
+<hidden 代码>
+<code cpp>
+#include<algorithm>
+#include<iostream>
+#include<cstdlib>
+#include<cstring>
+#include<cstdio>
+#include<vector>
+#include<queue>
+#include<cmath>
+#include<map>
+#include<set>
+#define LL long long int
+#define REP(i,n) for (int i = 1; i <= (n); i++)
+#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
+#define cls(s,v) memset(s,v,sizeof(s))
+#define mp(a,b) make_pair<int,int>(a,b)
+#define cp pair<int,int>
+using namespace std;
+const int maxn = 300005,maxm = 100005,INF = 0x3f3f3f3f;
+inline int read(){
+	int out = 0,flag = 1; char c = getchar();
+	while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
+	while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
+	return flag ? out : -out;
+}
+char A[maxn],B[maxn];
+int n,m;
+int main(){
+	char *a,*b; int tag;
+	while (~scanf("%s%s",A + 1,B + 1)){
+		tag = 0;
+		n = strlen(A + 1);
+		m = strlen(B + 1);
+		if (n > m) a = A,b = B;
+		else a = B,b = A,swap(n,m),tag = 1;
+		for (int i = 1; i <= n; i++) a[n + i] = a[i];
+		for (int i = 1; i <= m; i++){
+			for (int j = 1; i + j * m <= 2 * n; j++) b[i + j * m] = b[i];
+		}
+		int flag = 0;
+		for (int i = 1; i <= (n << 1); i++){
+			if (a[i] != b[i]){
+				if (a[i] > b[i]) flag = 1;
+				else if (a[i] < b[i]) flag = -1;
+				break;
+			}
+		}
+		if (flag == -1) puts(tag ? ">" : "<");
+		else if (flag == 1) puts(tag ? "<" : ">");
+		else puts("=");
+	}
+	return 0;
 }
 </code>
@@ 行 355: / 行 424: @@
 === 题解 ===
-solved by fyh hxm
+solve by fyh hxm
 先考虑无解的情况。$\lceil\frac{v_i}{u_i}\rceil>maxflow$是无解的，因为每一次都是流一次$\frac{u_i}{v_i}$，最多可以流$maxflow$次，如果$\frac{u_i}{v_i}*maxflow$还留不满1显然就是无解。
@@ 行 447: / 行 516: @@
 </code>
 </hidden>
-=====  =====
+===== J.Easy Integration =====
+=== 题意 ===
+算 $\int^{1}_{0}{(x-x^2)^ndx}$
+=== 题解 ===
+分部积分即可推出答案为 $\frac{(n!)^2}{(2n+1)!}$
+<hidden 代码>
+<code cpp>
+#include<iostream>
+#include<cstdio>
+#include<algorithm>
+#include<cstring>
+#define ll long long
+using namespace std;
+int read()
+{
+    int k=0,f=1;char c=getchar();
+    for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
+    for(;isdigit(c);c=getchar()) k=k*10+c-'0';return k*f;
+}
+const int N=2000005,M=1000000,mod=998244353;
+int n,frac[N];
+int ksm(int x,int k)
+{
+    int ans=1;
+    for(;k;k>>=1,x=1ll*x*x%mod)
+        if(k&1)
+            ans=1ll*ans*x%mod;
+    return ans;
+}
+int main()
+{
+    frac[1]=1;
+    for(int i=2;i<=M*2+1;i++)
+        frac[i]=1ll*i*frac[i-1]%mod;
+    while(scanf("%d",&n)!=EOF)
+    {
+        printf("%lld\n",1ll*frac[n]*frac[n]%mod*ksm(frac[2*n+1],mod-2)%mod);
+    }
+    return 0;
+}
+</code>
+</hidden>
 ===== 训练实况 =====

CVBB ACM Team

用户工具

站点工具

差别

页面工具