智算复赛
作者:互联网
解法:枚举
#include <bits/stdc++.h>
typedef long long ll ;
#define int ll
using namespace std ;
int quickpow(int a,int b,int mo){int s;for(s=1;b;b>>= 1,a=a*a%mo)if(b&1)s=s*a%mo;return s;}
#define INF 0x3f3f3f3f
const double PI = acos(-1.0);
const int maxn = 2e3+9;
const int mod = 998244353;
void Solve(){
int aa , b ;
scanf("%lld%lld" , &aa , &b);
if(b == 1){
puts("900");
}else if(b == 2){
if(aa%2) puts("0");
else puts("900");
}else{
int a = aa , len = 1 ;
while(a){
a /= 10 ;
len *= 10 ;
}
int cnt = 0 ;
for(int i = 1 ; i <= 9 ; i++){
for(int j = 0 ; j <= 9 ; j++){
for(int k = 0 ; k <= 9 ; k++){
int n = i*len*100+j*len*10+k*len+aa;
if(n%b==0){
cnt++;
}
}
}
}
printf("%lld\n" , cnt);
}
}
signed main(){
#ifdef ONLINE_JUDGE
#else
freopen("D:\\c++\\in.txt", "r", stdin);
//freopen("D:\\c++\\out.txt", "w", stdout);
#endif
//int t ;
//cin >> t ;
//while(t--)
Solve();
}
解法:一共要走2*n次,第一种方案每次走1,第二个方案每次走2,对比两种方案的性价比
如果第二种更优,则最大化选第二种,对坐标排序,类似最长子序列。
#include <bits/stdc++.h>
typedef long long ll ;
//#define int ll
using namespace std ;
int quickpow(int a,int b,int mo){int s;for(s=1;b;b>>= 1,a=a*a%mo)if(b&1)s=s*a%mo;return s;}
#define INF 0x3f3f3f3f
const double PI = acos(-1.0);
const int maxn = 2e3+9;
const int mod = 998244353;
int dp[maxn];
struct node{
int x , y ;
}a[maxn];
bool cmp(node a , node b){
if(a.x != b.x) return a.x < b.x;
return a.y < b.y ;
}
void Solve(){
int n , k , w1 , w2;
scanf("%d%d%d%d" , &n , &k , &w1 , &w2);
for(int i = 1 ; i <= k ; i++){
scanf("%d%d" , &a[i].x , &a[i].y);
dp[i] = 1 ;
}
if(2*w1 <= w2){
printf("%lld\n" , 2ll*n*w1);
}else{
sort(a+1 , a+1+k , cmp);
for(int i = 1 ; i <= k ; i++){
for(int j = 1 ; j < i ; j++){
if(a[i].x > a[j].x && a[i].y > a[j].y){
dp[i] = max(dp[i] , dp[j] + 1);
}
}
}
int ma = 0;
for(int i = 1 ; i <= k ; i++){
ma = max(ma , dp[i]);
}
printf("%lld\n" , 2ll*(n-ma)*w1 + 1ll*ma*w2);
}
}
signed main(){
#ifdef ONLINE_JUDGE
#else
freopen("D:\\c++\\in.txt", "r", stdin);
//freopen("D:\\c++\\out.txt", "w", stdout);
#endif
//int t ;
//cin >> t ;
//while(t--)
Solve();
}
解法:对于66个点,要构造出2的64次方的方案数,除去两个端点,根据二进制,这64个点每个点应该可以表示一位权值。
#include <bits/stdc++.h>
typedef unsigned long long ll ;
#define int ll
using namespace std ;
int quickpow(int a,int b,int mo){int s;for(s=1;b;b>>= 1,a=a*a%mo)if(b&1)s=s*a%mo;return s;}
#define INF 0x3f3f3f3f
const double PI = acos(-1.0);
const int maxn = 1e4+9;
const int mod = 998244353;
void Solve(){
int k , n ;
cin >> k >> n ;
if(k == 1){
printf("2 1\n1 2\n");
}else if(k == 2){
printf("3 3\n1 2\n2 3\n1 3\n");
}else if(k == 3){
printf("4 5\n1 2\n1 3\n3 2\n2 4\n1 4\n");
}else{
int cnt = 2016+64;
for(int i = 0 ; i <= 63 ; i++){
if((k>>i)&1){
cnt += 1 ;
}
}
printf("66 %lld\n" , cnt);
for(int i = 2 ; i <= 65 ; i++){
printf("1 %lld\n" , i);
for(int j = i+1 ; j <= 65 ; j++){
printf("%lld %lld\n" , i , j);
}
}
for(int i = 0 ; i <= 63 ; i++){
if((k>>i)&1){
printf("%lld 66\n" , i+2);
}
}
}
}
signed main(){
#ifdef ONLINE_JUDGE
#else
freopen("D:\\c++\\in.txt", "r", stdin);
//freopen("D:\\c++\\out.txt", "w", stdout);
#endif
//int t ;
//cin >> t ;
//while(t--)
Solve();
}
题意:求1-n的lcm
解法:每个只有一种素因子的数对答案有贡献,筛法时直接处理,注意数据范围要开无符号长整型。
欧拉筛更快,但注意会MLE,素数表长度不需要7e8.标记数组开bool
#include <bits/stdc++.h>
typedef long long ll ;
//#define int ll
int quickpow(int a,int b,int mo){int s;for(s=1;b;b>>= 1,a=a*a%mo)if(b&1)s=s*a%mo;return s;}
using namespace std ;
#define INF 0x3f3f3f3f
const double PI = acos(-1.0);
const int maxn = 8e7+1;
const ll mod = 4294967296 ;
int pr[5000000] , len;
bool is[maxn];
int vis[maxn];
void Solve(){
int n ;
ll a , b;
scanf("%d%lld%lld" , &n , &a , &b);
for(int i = 2 ; i <= n ; i++){
if(!is[i]){
pr[++len] = i ;
for(ll j = i ; j <= n ; j *= i){
vis[j] = i;
}
}
for(int j = 1 ; 1ll*i*pr[j] <= n ; j++){
is[i*pr[j]] = 1 ;
if(i%pr[j] == 0){
break;
}
}
}
for(int i = 1 ; i <= n ; i++){
if(vis[i]){
a = (a * vis[i] + b)%mod;
}
}
printf("%lld\n" , a);
}
signed main(){
#ifdef ONLINE_JUDGE
#else
freopen("D:\\c++\\in.txt", "r", stdin);
#endif
//int t ;
//scanf("%lld" , &t);
//while(t--)
Solve();
}
标签:const,int,mo,long,智算,ll,复赛,define 来源: https://www.cnblogs.com/nonames/p/13472393.html