牛客练习赛33
这题WildCow学长出的
C不想写结果最后中奖了２３３３
A:水题
#include
using namespace std;
typedef long long ll;
ll l,r,x;
ll F(ll y){
return y/x;
}
int main(){
int T;
cin>>T;
while(T--){
cin>>l>>r>>x;
cout<
B:一个线性筛，先统计一下，然后晒得时候乘个系数就行了，一个一个倍数标记肯定超时，比如输入的全是１的时候。
#include
using namespace std;
const int N=1e6+10;
int A[N],ANS[N];
int main(){
int n,k,x;
cin>>n>>k;
for(int i=1;i<=n;++i)scanf("%d",&x),A[x]++;
for(int i=1;i<=k;++i){
if(A[i]==0)continue;
for(int j=i;j<=k;j+=i){
ANS[j]+=A[i];
}
}
int ans=0;
for(int i=1;i<=k;++i){
ans=max(ans,ANS[i]);
}
int ans2=0;
for(int i=1;i<=k;++i){
if(ans==ANS[i])++ans2;
}
cout<
Ｃ:不想写，看最后三位能判断是否整除然后前边排序输出
D:一点小思维，把一个区间的一部分放到后边，涉及到的 数对 会取补，一开始看下逆序对的奇偶，连树状数组都不用，直接看置换环的奇偶，然后每次询问看涉及到的 数对 的奇偶
#include
using namespace std;
typedef long long ll;
const int N=1e5+10;
int A[N],B[N];
int main(){
int n;
scanf("%d",&n);
int ans=0;
for(int i=1;i<=n;++i)scanf("%d",&A[i]);
for(int i=1;i<=n;++i){
if(B[i]==0){
++ans;
B[i]=1;
int now=A[i];
while(B[now]==0){
B[now]=1;
now=A[now];
}
}
}
ans%=2;
int m;
scanf("%d",&m);
while(m--){
ll x,y,z;
scanf("%lld%lld%lld",&x,&y,&z);
z%=(y-x+1);
ll Ans=(y-x+1-z)*z;
if(Ans%2==1){
ans=!ans;
printf("%d\n",ans);
}else{
printf("%d\n",ans);
}
}
}
E:构造26个HASH函数，一个连续ｎ个１的HASH值，之后加减比较，这题一开始不知道脑子抽啥风，直接干了个线段树上去，ｍｄｚｚ。
#include
using namespace std;
typedef unsigned long long ll;
const int N=1e5+10;
char s[N];
ll F[N][26],P[N],G[N];
ll get(int l,int r,int x){
return F[r][x]-F[l-1][x]*P[r-l+1];
}
int main(){
int n,m;
scanf("%d",&n);
scanf("%s",s+1);
P[0]=1;
for(int i=1;i<=n;++i){
for(int j=0;j<=s[i]-'a';++j){
F[i][j]=F[i-1][j]*131+s[i]-'a';
}
for(int j=s[i]-'a';j<26;++j){
F[i][j]=F[i-1][j]*131+s[i]-'a'+26;
}
P[i]=P[i-1]*131;
}
for(int i=1;i<=n;++i){
G[i]=G[i-1]*131+1;
}
scanf("%d",&m);
while(m--){
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
int a=s[x]-'a';
int b=s[y]-'a';
if(a>b){
if(get(x,x+z-1,a)==get(y,y+z-1,b)+G[z]*(a-b))printf("%d\n",min(a-b,26-a+b));
else printf("-1\n");
}else{
if(get(x,x+z-1,a)+G[z]*(b-a)==get(y,y+z-1,b))printf("%d\n",min(b-a,26-b+a));
else printf("-1\n");
}
}
}
F:纸上找找规律会发现就是Ａ[n][n]的值，然后有个长度为ｎ的式子，一直卡在这个式子怎么算，后来打了下表发现就是阶乘，然后搜了个大数阶乘取模的板子，原理是分段打表：
#include
#include
#define lon long long
using namespace std;
lon A[1010],B[1010];
lon n,p;
lon a[110]={1,682498929,491101308,76479948,723816384,67347853,27368307,
625544428,199888908,888050723,927880474,281863274,661224977,623534362,
970055531,261384175,195888993,66404266,547665832,109838563,933245637,
724691727,368925948,268838846,136026497,112390913,135498044,217544623,
419363534,500780548,668123525,128487469,30977140,522049725,309058615,
386027524,189239124,148528617,940567523,917084264,429277690,996164327,
358655417,568392357,780072518,462639908,275105629,909210595,99199382,
703397904,733333339,97830135,608823837,256141983,141827977,696628828,
637939935,811575797,848924691,131772368,724464507,272814771,326159309,
456152084,903466878,92255682,769795511,373745190,606241871,825871994,
957939114,435887178,852304035,663307737,375297772,217598709,624148346,
671734977,624500515,748510389,203191898,423951674,629786193,672850561,
814362881,823845496,116667533,256473217,627655552,245795606,586445753,
172114298,193781724,778983779,83868974,315103615,965785236,492741665,
377329025,847549272,698611116};
int main()
{
int m;
p=1000000007;
cin>>n>>m;
for(int i=1;i<=m;++i)cin>>A[i]>>B[i];
lon ANS;
for(int i=1;i<=m;++i){
if(B[i]==n){
ANS=A[i];
}
}
lon now=n/10000000;
lon ans=a[now];
for(lon i=now*10000000+1;i<=n;i++)
ans=ans*i%p;
cout<