Codeforces Round #365 (Div. 2) 题解
2016-08-21 13:28
302 查看
Mishka and Game
Mishka and trip
Chris and Road
Mishka and Interesting sum
Mishka and Divisors
1 000 000次询问，
求一个区间出现偶数(且>=2)次的数的xor和。
首先区间xor和与区间和都能用树状数组解决
奇数的xor和很好求，考虑求一个区间出现过的数的xor和
我们维护一个数列，保证到第i个元素为止最后一次出现数的位置上才有数。
所以不用滚动数组，判断mp(f[i][h[k]],s[i][h[k]])==mp(f[i−1][h[k]],s[i−1][h[k]]),f是背包，s是最优值
Mishka and trip
Chris and Road
Mishka and Interesting sum
Mishka and Divisors
Mishka and Game
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a
<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
int main()
{
// freopen("A.in","r",stdin);
// freopen(".out","w",stdout);
int n=read();
int a=0,b=0;
For(i,n) {
int u=read(),v=read();
a+=u>v; b+=v>u;
}
if (a==b) puts("Friendship is magic!^^"); else
if (a<b) puts("Chris"); else
puts("Mishka");
return 0;
}Mishka and trip
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a
<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
#define MAXN (100000+10)
ll c[MAXN],s[MAXN]={0},t[MAXN]={0};
bool b[MAXN]={0};
int main()
{
// freopen("B.in","r",stdin);
// freopen(".out","w",stdout);
int n=read(),k=read();
For(i,n) c[i]=read();
For(i,n) s[i]=s[i-1]+c[i];
For(i,k) b[read()]=1;
For(i,n) t[i]=t[i-1]+b[i]*c[i];
ll ans=0;
For(i,n) {
if (b[i]) ans+=c[i]*(s
-s[i]);
else {
if (i==1) ans+=c[1]*(c[2]+c
+t[n-1]-t[2]);
else if (i<n) ans+=c[i]*(c[i+1]+t
-t[i+1]);
}
}
cout<<ans<<endl;
return 0;
}Chris and Road
注意精度#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a
<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
int n;
double w,u,v;
int main()
{
// freopen("C.in","r",stdin);
// freopen(".out","w",stdout);
cin>>n>>w>>v>>u;
vector<double> q;
For(i,n) {
double x=read();int y=read();
if (y>0) x-=(double)y/u*v;
q.pb(x);
}
sort(q.begin(),q.end());
if (q[0]>=-1e-12||q[n-1]<=1e-12) printf("%.10lf\n",w/u);
else {
printf("%.10lf\n",q[n-1]/v+w/u);
}
return 0;
}Mishka and Interesting sum
给一个 1 000 000长度的数列，1 000 000次询问，
求一个区间出现偶数(且>=2)次的数的xor和。
首先区间xor和与区间和都能用树状数组解决
奇数的xor和很好求，考虑求一个区间出现过的数的xor和
我们维护一个数列，保证到第i个元素为止最后一次出现数的位置上才有数。
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a
<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
int n,m;
#define MAXN (1000100)
ll a[MAXN]={0},s[MAXN]={0};
ll ans[MAXN]={0};
ll f[MAXN]={0};
void upd(int x,ll v){for(;x<=n;x+=x&(-x)) f[x]^=v;}
ll get(ll x){ll v=0; for(;x;x-=x&(-x)) v^=f[x]; return v;}
vi rr[MAXN];
int l[MAXN];
map<int, int> ls;
int main()
{
// freopen("D.in","r",stdin);
// freopen(".out","w",stdout);
cin>>n;
For(i,n) {
a[i]=read(); ls[a[i]]=0;
s[i]=a[i]^s[i-1];
}
cin>>m;
For(i,m) {
int r;
l[i]=read(); r=read();
ans[i]=s[r]^s[l[i]-1];
rr[r].pb(i);
}
For(i,n) {
upd(i,a[i]);
if (ls[a[i]]) upd(ls[a[i]],a[i]);
ls[a[i]]=i;
Rep(k,SI(rr[i])) {
int j=rr[i][k];
ans[j]^=get(i)^get(l[j]-1);
}
}
For(i,m) {
printf("%I64d\n",ans[i]);
}
return 0;
}Mishka and Divisors
普通的01背包，但是要输出最优解时取了哪些元素，状态难以保存，所以不用滚动数组，判断mp(f[i][h[k]],s[i][h[k]])==mp(f[i−1][h[k]],s[i−1][h[k]]),f是背包，s是最优值
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a
<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
#define MAXN (15000)
vector<ll> p;
map<ll,int> h;
int sizh=0;
ll f[1010][MAXN],s[1010][MAXN];
ll a[MAXN],a2[MAXN];
ll gcd(ll a,ll b){if (!b) return a;return gcd(b,a%b);}
int main()
{
// freopen("E.in","r",stdin);
// freopen(".out","w",stdout);
int n=read();
ll k;
cin>>k;
for(ll i=1;i*i<=k;i++) {
if (k%i==0) {
p.pb(i);
if (i*i!=k) p.pb(k/i);
}
}
sort(p.begin(),p.end());
int sz=p.size();
Rep(i,sz) h[p[i]]=i,f[0][i]=s[0][i]=INF;
For(i,n) {
cin>>a[i];
a2[i]=a[i];
a[i]=gcd(a[i],k);
}
f[0][h[1]]=0; s[0][h[1]]=0;
For(i,n) {
RepD(j,sz-1){
ll l=p[j];
ll t=l/gcd(l,a[i]);
int hl=h[l],ht=h[t];
f[i][hl]=f[i-1][hl]; s[i][hl]=s[i-1][hl];
ll fl=f[i-1][hl],ft=f[i-1][ht],sl=s[i-1][hl],st=s[i-1][ht];
if (fl>ft+1) {
f[i][hl]=ft+1;
s[i][hl]=st+a2[i];
}
else if (fl==ft+1&&sl>st+a2[i])
{
s[i][hl]=st+a2[i];
}
}
}
if (f
[h[k]]==INF) puts("-1") ;
else {
if (k==1) {
puts("1");
printf("%d\n",(int)(min_element(a2+1,a2+1+n)-a2));
return 0;
}
printf("%I64d\n",f
[h[k]]);
ForD(i,n) {
if (mp(f[i][h[k]],s[i][h[k]])==mp(f[i-1][h[k]],s[i-1][h[k]])) continue;
k/=gcd(k,a[i]);
printf("%d ",i);
}
cout<<endl;
}
return 0;
}
内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理 
相关文章推荐
- 树形DP______Rebuilding Roads( POJ 1947 )
- python 对图片的截取
- #187 Gas Station
- Qt 应用程序的部署
- 2016年将已有.db数据库添加到ios项目里的方法.
- 经典算法学习——在O(1)时间删除链表节点
- Laravel 安装多国语言包后，phpstorm 还是报错
- 《数据结构与算法》学习笔记14 用链表实现抽象数据类型
- React 学习笔记
- Ugly Number（丑数）
- java分页
- Codeforces Round #368 (Div. 2) Pythagorean Triples
- 使用Maven创建项目时总是出现异常！maven-archetype-quickstart1.1有问题
- Codeforces Round #368 (Div. 2) A(水题) B(枚举 思维) C(数论 勾股数)
- 模拟实现String类(2)——写时拷贝
- 使用svnsync同步svn
- 转：Intent与PendingIntent
- 原生JS实现-星级评分系统的简单实例
- [Android 之美] 那些你不知道的APK 瘦身，让你的APK更小
- python 各种装饰器示例（python3）