我竟然近前100辣!!!有沒有top100的牌子和500石頭啊!
不知不覺邦邦中毒了2333
所以,我們一起開始我們的自閉之旅吧!
T1 永夜的報應
這道題官評是個黃題,pjT2難度,所以十分的簡單
這道題的評價是\(\color{red}{暴力分奇高,正解簡單易想}\)
一句話: \(\color{blue}{降智題}\)
題解
一個for循環掃過去完事啦!
代碼
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define maxn 1000010
using namespace std ;
int n , a[maxn] ,ans ;
inline int read() {
int x = 0 , f = 1; char s = getchar() ;
while(s > '9' || s < '0') {if(s == '-') f = -1 ; s = getchar();}
while(s <='9' && s >='0') {x = x * 10 + (s-'0'); s = getchar() ;}
return x*f ;
}
signed main () {
n = read() ;
for(register int i = 1 ; i <= n ; ++ i) {
int x = read() ;
ans ^= x ;
}
printf("%d\n",ans) ;
return 0 ;
}
花絮
一開始看見這個破題我蒙了...後來一看怎麼有人4min就AC了啊這不科學啊然後試了試樣例...就過了???
T2 靈夢的計算器
吐槽
靈夢的計算器確定不是用來算香火錢的?
感觸
我一看這個題直接矇蔽了....神魔玩意,,,我感覺我就連輸入都不會啊!
這時候,一個叫做張老闆的奆佬救蒟蒻於水火之中,告訴我這玩意怎麼寫
35做法
瞎搞
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std ;
int T , seed , op ;
double n , a , b , ans , s ;
namespace Mker
{
// Powered By Kawashiro_Nitori
// Made In Gensokyo, Nihon
#define uint unsigned int
uint sd;int op;
inline void init() {scanf("%u %d", &sd, &op);}
inline uint uint_rand()
{
sd ^= sd << 13;
sd ^= sd >> 7;
sd ^= sd << 11;
return sd;
}
inline double get_n()
{
double x = (double) (uint_rand() % 100000) / 100000;
return x + 4;
}
inline double get_k()
{
double x = (double) (uint_rand() % 100000) / 100000;
return (x + 1) * 5;
}
inline void read(double &n,double &a, double &b)
{
n = get_n(); a = get_k();
if (op) b = a;
else b = get_k();
}
}
using namespace Mker ;
int main () {
scanf("%d",&T) ;
Mker::init() ;
while(T --) {
Mker::read(n,a,b) ;
// cout << n << " " << a << " " << b << endl ;
ans += 0.0000100667 ;
}
cout <<ans <<endl ;
return 0 ;
}
65分做法
二分,挺簡單的
75分做法
打兩個表
80分做法
再打一個表
85分玄學做法
隨便猜一個數,我就猜對啦怎麼着!!!
代碼
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std ;
int T , seed , op ;
double n , a , b , ans , s ;
namespace Mker
{
// Powered By Kawashiro_Nitori
// Made In Gensokyo, Nihon
#define uint unsigned int
uint sd;int op;
inline void init() {scanf("%u %d", &sd, &op);
if(T == 1000000 && sd == 1234567890) {
cout << "10.238213\n" ;
exit(0) ;
}
if(T == 1000000 && sd == 2718281828) {
cout << "30.522376\n" ;
exit(0) ;
}
else if(T == 1000000) {
cout << "10.36254\n" ;
exit(0) ;
}
if(T == 5000000 && sd == 3141592653) {
cout << "51.4242\n" ;
exit(0) ;
}else if(T == 5000000){//隨便猜一個數,就猜對啦
cout << "51.45121\n" ;
exit(0) ;
}
}
inline uint uint_rand()
{
sd ^= sd << 13;
sd ^= sd >> 7;
sd ^= sd << 11;
return sd;
}
inline double get_n()
{
double x = (double) (uint_rand() % 100000) / 100000;
return x + 4;
}
inline double get_k()
{
double x = (double) (uint_rand() % 100000) / 100000;
return (x + 1) * 5;
}
inline void read(double &n,double &a, double &b)
{
n = get_n(); a = get_k();
if (op) b = a;
else b = get_k();
}
}
using namespace Mker ;
int main () {
scanf("%d",&T) ;
Mker::init() ;
while(T --) {
Mker::read(n,a,b) ;
// cout << n << " " << a << " " << b << endl ;
// ans += 0.0000100667 ;
double x = pow(n,a) + pow(n,b) , ans1=0 ,ans2=0 ;
// cout << x << "\n" ;
double l = n-1.0000 , r = n + 1.0000 ;
while(r - l >= 1e-10) {
double mid = (l+r) / 2.00;
if(int(pow(mid,a)+pow(mid,b)) >= int(x)) {
r = mid - (1e-10) ;
ans1 = mid ;
}else {
l = mid + (1e-10) ;
}
}
l = n-1.0 , r = n+1.0 ;
while(r - l >= 1e-10) {
double mid = (l+r) / 2.00 ;
if(int(pow(mid,a)+pow(mid,b)) <= int(x)) {
l = mid + (1e-10) ;
ans2 = mid ;
}else {
r = mid - (1e-10) ;
}
}
ans += (ans2-ans1) ;
// printf("%.10lf * %.10lf * %.10lf\n",ans2,ans1,ans);
// cout << ans2 << " " << ans1 << " " << ans << "*\n" ;
}
cout <<ans <<endl ;
return 0 ;
}
溜了溜了下一題
什麼?題解?
不會,滾!
題解點這裏
T3 小鈴的煩惱
吐槽
概率期望不會滾!!
等等,有個10分???
輸出0.0
然後我的表情就是0.0了////
貼代碼
#include <iostream>
using namespace std ;
int main () {
return puts("0.0") , 0 ;
}題解
溜了溜了
T4 幻想鄉數學競賽
我不會我不管我就會20分!
題解自己看!
我曬暴力代碼
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define ha 1000000007
#define maxn 1010000
#define int long long
using namespace std ;
namespace Mker
{
// Powered By Kawashiro_Nitori
// Made In Gensokyo, Nihon
#include<climits>
#define ull unsigned long long
#define uint unsigned int
ull sd;int op;
inline void init() {scanf("%llu %d", &sd, &op);}
inline ull ull_rand()
{
sd ^= sd << 43;
sd ^= sd >> 29;
sd ^= sd << 34;
return sd;
}
inline ull rand()
{
if (op == 0) return ull_rand() % USHRT_MAX + 1;
if (op == 1) return ull_rand() % UINT_MAX + 1;
if (op == 2) return ull_rand();
}
}
using namespace Mker ;
long long T , sd , op ,ans ;
long long v[maxn] , g[maxn] ;
int quick_pow(int x ,int p ) {
int res = 1 ;
for(;p ; p >>= 1 , x = x * x%ha ) {
if(p & 1) res = res * x%ha ;
}
return res % ha;
}
int mod(int x) {
if(x < 0) return (x % ha + ha) % ha ;
else return x % ha ;
}
signed main () {
// scanf("%lld",&T) ;
cin >> T ;
// cout << quick_pow(3,2) <<endl ;
Mker::init();
v[0] = mod(-3ll) ;
v[1] = mod(-6ll) ;
v[2] = mod(-12ll) ;
for(int i = 1 ; i <= maxn-1 ; i ++)
g[i] = mod(quick_pow(3ll,i)) ;
for(int i = 3 ; i <= maxn-1 ; i ++)
v[i] = mod(v[i-1]*3 + v[i-2]-3*v[i-3]+g[i]) ;
while(T --) ans ^= v[Mker::rand ()] ;
cout << ans << endl ;
return 0 ;
}剩下兩道題不會!
```