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+7; const int maxn = 1e2+5; const double eps = 1e-6; 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==0?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=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';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,1,n+1){ d[i][1] = i - 1; rep(j,2,n+1){ d[i][j] = d[i][j-1] + 1; } } } int main(){ scanf("%d",&n); nn = n; init(); if(n==2) return printf("3 4 5 2 3 "),0; printf("%d",n); rep(i,1,n+1) rep(j,1,n+1) if(d[i][j]>n) printf(" %d",(i-1)*n+j),d[i][j]=-1; puts(""); n = (n % 2?n + 2:n + 1); int dis = nn; rep(i,0,nn){ printf("%d",n); rep(j,1,nn+1) rep(k,1,nn+1) if(d[j][k]==dis) printf(" %d",(j-1)*nn+k); puts("");dis--;n += 2; } return 0; }
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+7; const int maxn = 6e3+5; const double eps = 1e-6; 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==0?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=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';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"),0; int ss=0,tt=0; rep(i,0,l) { if(s[i]=='+') a[ss++]=i; if(t[i]=='+') b[tt++]=i; } if(ss!=tt) return puts("-1"),0; int ans = 0; rep(i,0,ss) ans += abs(a[i]-b[i]); cout<<ans<<endl; return 0; }
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+7; const int maxn = 1e5+5; const double eps = 1e-6; 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==0?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=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';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 = 0; rep(i,0,n){ scanf("%d",&a); if(pow(a,m)>0) ans += pow(a,m); } printf("%d ",ans); return 0; }
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+7; const int maxn = 16005; const double eps = 1e-6; 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==0?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=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';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,0,sz(G[u])){ int v=G[u][i]; if(v!=p){ int tmp = dfs(v,u); if(tmp>0) ret+=tmp;; } } // dd(u)de(dp[u]) return ret; } int main(){ int n,u,v; scanf("%d",&n); rep(i,1,n+1) dp[i]=-inf; rep(i,1,n+1) scanf("%d",a+i); rep(i,1,n) scanf("%d%d",&u,&v),add(u,v); dfs(1,-1); int ans = -inf; rep(i,1,n+1) ans=max(ans,dp[i]); printf("%d ",ans); return 0; }