https://codingcompetitions.withgoogle.com/kickstart/round/0000000000050e02/000000000018fd0d
Book Reading (10pts, 15pts)
1 ≤ T ≤ 100.
1 ≤ P1 < P2 < ... < PM ≤ N.
1 ≤ Ri ≤ N, for all i.
Test set 1 (Visible)
1 ≤ M ≤ N ≤ 1000.
1 ≤ Q ≤ 1000.
Test set 2 (Hidden)
1 ≤ M ≤ N ≤ 105.
1 ≤ Q ≤ 105.
题意：一本书有 1……n页，撕去 p 页，有 q 个读者， 问读者一共能阅读的页数
如果直接枚举暴力的话， 10ps 到手
#include<bits/stdc++.h>
using namespace std;
#define _for(i,a,b) for(int i = (a); i < (b); i++)
#define _rep(i,a,b) for(int i = (a); i <= (b); i++)
const int N = 1e5+100;
int a[N],b[N],vis[N];
int main(){
ios::sync_with_stdio(0); cin.tie(0);
int t,n,m,q,kase = 1; cin >>t ;
while(t--) {
cin >> n >> m >> q; int x,cnt = 0;
memset(vis, 0, sizeof(vis));
_for(i,0,m) cin >> x, vis[x] = 3;
_for(i,0,q) {
cin >> x;
for(int j = x; j <= n; j += x)
if(vis[j] != 3) cnt++;
}
cout << "Case #"<< kase++ << ": ";
cout << cnt << endl;
}
return 0;
}
但是很明显， 会 TLE， 所以需要优化下
#include<bits/stdc++.h>
using namespace std;
#define _for(i,a,b) for(int i = (a); i < (b); i++)
#define _rep(i,a,b) for(int i = (a); i <= (b); i++)
#define ll long long
const int N = 1e5+100;
int main(){
ios::sync_with_stdio(0); cin.tie(0);
int t,n,m,q,kase = 1; cin >>t ;
while(t--) {
vector<int> a(N, 0);
vector<bool> vis(N, false);
cin >> n >> m >> q; int x;
ll cnt = 0;
_for(i,0,m) cin >> x, vis[x] = true;
_for(i,0,q) {
cin >> x;
if(a[x]){
cnt += a[x];
continue;
}
for(int j = x; j <= n; j += x)
if(!vis[j]) a[x]++;
cnt += a[x];
}
cout << "Case #"<< kase++ << ": ";
cout << cnt << endl;
}
return 0;
}
https://codingcompetitions.withgoogle.com/kickstart/round/0000000000050e02/000000000018fe36
The Equation (12pts, 20pts)
直接暴力 12pts可到手
#include<bits/stdc++.h>
using namespace std;
const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;
#define scld(x) scanf("%lld", &x)
#define prcas printf("Case #%d: ", cas)
#define pncas printf("Case #%d:\n", cas)
#define lowbit(x) ((x) & (-(x)))
#define fi first
#define se second
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef vector<int> vi;
const int maxn = 25;
int dp[1048550];
LL a[maxn], b[maxn];
int n, T, h;
int main()
{
scanf("%d", &T);
for(int cas = 1; cas <= T; ++ cas)
{
scanf("%d %d", &n, &h);
for(int i = 0; i < n; ++ i) scld(a[i]);
//for(int i = 0; i < n; ++ i) scld(b[i]);
prcas; int flag = 0;
for(int i = 128; i >= 0; i--){
int sum = 0;
for(int j = 0; j < n; j++)
sum += a[j]^i;
if(sum <= h) {
flag = 1; printf("%d\n", i);break;
}
}
if(!flag) printf("-1\n");
}
return 0;
}
通过对 k 的二进制位数进行枚举， cnt[i][0] cnt[i][1]分别记录 数组a的 第i+1位的 0 与 1的个数
再通过 ret[i][0] ret[i][1] 分别表示对前 i 位的a数组的 0 与 1 表示的和
然后 通过 suf[i] 记录前 i 项 min(ret[i][0], ret[i][1])的和， （？？？）
然后枚举 位数 i 51~0 如果符合 ret[i][1]+suf[i-1] k |= 1<<i , 即 k 的这个二进制位为 1， 否则为 0 m-= ( 为1 ? ret[i][1] : ret[i][0])
若枚举结束 m < 0， 则不存在 k
#include<bits/stdc++.h>
using namespace std;
const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
#ifdef __LOCAL_DEBUG__
# define _debug(fmt, ...) fprintf(stderr, "[%s] " fmt "\n", __func__, ##__VA_ARGS__)
#else
# define _debug(...) ((void) 0)
#endif
#define IN freopen("B.in", "r", stdin)
#define OUT freopen("B.out", "w", stdout)
#define scd(x) scanf("%d", &x)
#define scld(x) scanf("%lld", &x)
#define scs(x) scanf("%s", x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define prcas printf("Case #%d: ", cas)
#define pncas printf("Case #%d:\n", cas)
#define lowbit(x) ((x) & (-(x)))
#define fi first
#define se second
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef vector<int> vi;
const int maxk = 51;
int cnt[65][2];
int n, T;
LL m, ans, ret[65][2], suf[65];
int main()
{
scd(T);
for(int cas = 1; cas <= T; ++ cas)
{
for(int i = 0; i <= maxk; ++ i) cnt[i][0] = cnt[i][1] = 0;
scanf("%d%lld", &n, &m);
while(n --)
{
LL x; scanf("%lld", &x); ans = 0;
for(int i = 0; i <= maxk; ++ i, x >>= 1) cnt[i][x & 1] ++;
}
for(int i = 0; i <= maxk; ++ i)
ret[i][0] = 1ll * cnt[i][1] * (1ll << i), ret[i][1] = 1ll * cnt[i][0] * (1ll << i);
suf[0] = min(ret[0][0], ret[0][1]);
for(int i = 1; i <= maxk; ++ i)
suf[i] = suf[i - 1] + min(ret[i][0], ret[i][1]);
for(int i = maxk; i >= 0 && m >= 0; -- i)
{
LL t = 0; if(i > 0) t = suf[i - 1];
if(ret[i][1] + t <= m)
{
ans |= 1ll << i;
m -= ret[i][1];
}else m -= ret[i][0];
}
if(m < 0) ans = -1;
prcas; printf("%lld\n", ans);
}
return 0;
}
/*
Input
4
3 27
8 2 4
4 45
30 0 4 11
1 0
100
6 2
5 5 1 5 1 0
Case #1: 12
Case #2: 14
Case #3: 100
Case #4: -1
*/
https://codingcompetitions.withgoogle.com/kickstart/round/0000000000050e02/000000000018fd5e
Shifts (20pts, 23pts)
题意：有n个班次， h 快乐值要求， 问有多少种情况 a 与 b 都快乐（快乐值大于等于h）
如果直接暴力的话， 0~ 3^n-1, 暴力枚举 每个班次的三种情况（单选a， 单选b， 同时选ab），n<=12，5e5的样子，能过 Test set 1， 但Test set 2 3e8（n<=20）就会TLE
TLE
#include<bits/stdc++.h>
using namespace std;
#define rep(x,y) for(long long i =x;i<=y;i++)
#define pb push_back
#define mp make_pair
#define ALL(x) (x).begin(),(x).end()
#define FastIO ios_base::sync_with_stdio(false); cin.tie(NULL);
typedef long long ll;
typedef pair< ll, ll> pll;
typedef vector< ll > vll;
typedef map < ll, ll > mll;
int main()
{
ll i,j,t,test,ans,n,num,s1,s2,a[30],b[30],h,ways;
test=1;
cin >> t;
while(t--)
{
cout << "Case #" << test << ": " ;
ans=0;
test++;
cin >> n >> h;
rep(1,n)
cin >> a[i];
rep(1,n)
cin >> b[i];
ways= pow(3,n)-1;
rep(0,ways)
{
num=i;
j=1;
s1=0;
s2=0;
while(j<=n)
{
if(num%3==0)
{
s1+=a[j];
}
else if(num%3==1)
s2+=b[j];
else
{
s1+=a[j];
s2+=b[j];
}
num/=3;
j++;
}
// cout << s1 << " " << s2 << endl;
if(s1>=h && s2>=h) ans++;
}
cout << ans << endl;
}
return 0;
}
dp[mask] mask 0~2^n-1, 以0 1 分 n个a是否被取
then mask 0~2^n-1 i 0~n dp[mask^(1<<i)] += dp[mask]
then x 0~2^n-1 dp[x] 2^n-x-1(二进制01 代表n个b是否被取) 神奇的穷举操作
最后会发现 所有的可能 3^n 都被枚举了
比如说 n = 2,
a b
dp[0] = 00 01 10 11 11
dp[1] = 01 11 10
dp[2] = 10 11 01
dp[3] = 11 00
#include<bits/stdc++.h>
using namespace std;
const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;
#define scld(x) scanf("%lld", &x)
#define prcas printf("Case #%d: ", cas)
#define pncas printf("Case #%d:\n", cas)
#define lowbit(x) ((x) & (-(x)))
#define fi first
#define se second
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef vector<int> vi;
const int maxn = 25;
int dp[1048550];
LL a[maxn], b[maxn];
int n, T, h;
int main()
{
scanf("%d", &T);
for(int cas = 1; cas <= T; ++ cas)
{
scanf("%d %d", &n, &h);
for(int i = 0; i < n; ++ i) scld(a[i]);
for(int i = 0; i < n; ++ i) scld(b[i]);
for(int mask = 0; mask < (1 << n); ++ mask)
{
LL sum = 0;
for(int i = 0; i < n; ++ i)
if((mask >> i) & 1) sum += a[i];
if(sum >= h) dp[mask] = 1; else dp[mask] = 0;
}
//for(int mask = 0; mask < (1 << n); ++ mask) printf("dp[%d]= %d ", mask, dp[mask]);printf("\n");
for(int i = 0; i < n; ++ i)
for(int mask = 0; mask < (1 << n); ++ mask)
if((mask >> i) & 1) dp[mask ^ (1 << i)] += dp[mask];
LL ans = 0;
//for(int mask = 0; mask < (1 << n); ++ mask) printf("dp[%d]= %d ", mask, dp[mask]);printf(" \t...\n");
for(int mask = 0; mask < (1 << n); ++ mask)
{
LL sum = 0, task = 0;
for(int i = 0; i < n; ++ i)
if((mask >> i) & 1) sum += b[i];
else task |= (1 << i);
if(sum < h) continue;
ans += dp[task];
}
prcas; printf("%lld\n", ans);
}
return 0;
}
原文地址：https://www.cnblogs.com/163467wyj/p/11728925.html
时间： 2024-11-03 03:13:22