今日SGU 5.8

SGU 109

题意：一个n*n的矩形，起点在1,1然后每次给你一个操作，走ki步，然后你可以删除任意一个点这次步走不到的，删了就不能再走了，然后问构造这种操作，使得最后删除n*n-1个点

剩下一个点，这个人最终的目的就在那，还要求每次走的步数要递增，n<=ki<300

收获：奇妙的构造，每次走奇数点，就会走到和自己奇偶不一样的点，（1,1）为偶点，（2,3）为奇点，然后第一次先删除n步走不到的点，那么接下来的点到起点的距离就是0-n，那么我们就可以

删n次，然后每一次删距离为n-i的点，最终他就会被逼到（1,1）了，太秀了

#include<bits/stdc++.h>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
#define ll long long
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define pii pair<int,int>
#define pdd pair<double,double>
#define pdi pair<double,int>
#define mp(u,v) make_pair(u,v)
#define sz(a) (int)a.size()
#define ull unsigned long long
#define ll long long
#define pb push_back
#define PI acos(-1.0)
#define qc std::ios::sync_with_stdio(false)
#define db double
#define all(a) a.begin(),a.end()
const int mod = 1e9+;
const int maxn = 1e2+;
const double eps = 1e-;
using namespace std;
bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
bool ls(const db &a, const db &b) { return a + eps < b; }
bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=,f=;char ch=getchar();
    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//inv[1]=1;
//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
int n,nn;
int d[maxn][maxn];
void init(){
    rep(i,,n+){
        d[i][] = i - ;
        rep(j,,n+){
            d[i][j] = d[i][j-] + ;
        }
    }
}
int main(){
    scanf("%d",&n);
    nn = n;
    init();
    if(n==) return printf("3 4\n5 2 3\n"),;
    printf("%d",n);
    rep(i,,n+) rep(j,,n+) if(d[i][j]>n) printf(" %d",(i-)*n+j),d[i][j]=-;
    puts("");
    n = (n % ?n + :n + );
    int dis = nn;
    rep(i,,nn){
        printf("%d",n);
        rep(j,,nn+) rep(k,,nn+) if(d[j][k]==dis) printf(" %d",(j-)*nn+k);
        puts("");dis--;n += ;
    }
    return ;
}

SGU 170

题意：可以移动相邻的+,-然后问你把两个字符串变成一样的最少次数

收获：就是贪心，每个+找最近的可以匹配的

#include<bits/stdc++.h>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
#define ll long long
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define pii pair<int,int>
#define pdd pair<double,double>
#define pdi pair<double,int>
#define mp(u,v) make_pair(u,v)
#define sz(a) (int)a.size()
#define ull unsigned long long
#define ll long long
#define pb push_back
#define PI acos(-1.0)
#define qc std::ios::sync_with_stdio(false)
#define db double
#define all(a) a.begin(),a.end()
const int mod = 1e9+;
const int maxn = 6e3+;
const double eps = 1e-;
using namespace std;
bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
bool ls(const db &a, const db &b) { return a + eps < b; }
bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=,f=;char ch=getchar();
    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//inv[1]=1;
//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
char s[maxn],t[maxn];
int a[maxn],b[maxn];
int main(){
    scanf("%s%s",s,t);
    int l = strlen(s),l2 = strlen(t);
    if(l!=l2) return puts("-1"),;
    int ss=,tt=;
    rep(i,,l) {
        if(s[i]=='+') a[ss++]=i;
        if(t[i]=='+') b[tt++]=i;
    }
    if(ss!=tt) return puts("-1"),;
    int ans = ;
    rep(i,,ss) ans += abs(a[i]-b[i]);
    cout<<ans<<endl;
    return ;
}

SGU 163

题意：给你n个数，数字范围-3到3，在给你一个m，问你怎么选这些数字，可选可不选，使得这些数字的m次幂和最大

收获：无

#include<bits/stdc++.h>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
#define ll long long
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define pii pair<int,int>
#define pdd pair<double,double>
#define pdi pair<double,int>
#define mp(u,v) make_pair(u,v)
#define sz(a) (int)a.size()
#define ull unsigned long long
#define ll long long
#define pb push_back
#define PI acos(-1.0)
#define qc std::ios::sync_with_stdio(false)
#define db double
#define all(a) a.begin(),a.end()
const int mod = 1e9+;
const int maxn = 1e5+;
const double eps = 1e-;
using namespace std;
bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
bool ls(const db &a, const db &b) { return a + eps < b; }
bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=,f=;char ch=getchar();
    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//inv[1]=1;
//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
int main(){
    int n,m,a;
    scanf("%d%d",&n,&m);
    int ans = ;
    rep(i,,n){
        scanf("%d",&a);
        if(pow(a,m)>) ans += pow(a,m);
    }
    printf("%d\n",ans);
    return ;
}

SGU 143

题意：给你一颗树，每个点有权值，要你从中选出一颗子树，使得子树的权值最大

收获：树上dp

#include<bits/stdc++.h>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
#define ll long long
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define pii pair<int,int>
#define pdd pair<double,double>
#define pdi pair<double,int>
#define mp(u,v) make_pair(u,v)
#define sz(a) (int)a.size()
#define ull unsigned long long
#define ll long long
#define pb push_back
#define PI acos(-1.0)
#define qc std::ios::sync_with_stdio(false)
#define db double
#define all(a) a.begin(),a.end()
const int mod = 1e9+;
const int maxn = ;
const double eps = 1e-;
using namespace std;
bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
bool ls(const db &a, const db &b) { return a + eps < b; }
bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=,f=;char ch=getchar();
    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//inv[1]=1;
//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
int a[maxn],dp[maxn];
vector<int> G[maxn];
void add(int u,int v){
    G[u].pb(v);G[v].pb(u);
}
int dfs(int u,int p){
    int& ret = dp[u];
    ret = a[u];
    rep(i,,sz(G[u])){
        int v=G[u][i];
        if(v!=p){
            int tmp = dfs(v,u);
            if(tmp>) ret+=tmp;;
        }
    }
//    dd(u)de(dp[u])
    return ret;
}
int main(){
    int n,u,v;
    scanf("%d",&n);
    rep(i,,n+) dp[i]=-inf;
    rep(i,,n+) scanf("%d",a+i);
    rep(i,,n) scanf("%d%d",&u,&v),add(u,v);
    dfs(,-);
    int ans = -inf;
    rep(i,,n+) ans=max(ans,dp[i]);
    printf("%d\n",ans);
    return ;
}