1. 815A Karen and Game
大意: 给定$nm$矩阵, 每次选择一行或一列全部减$1$, 求最少次数使得矩阵全$0$
贪心, $n>m$时每次取一列, 否则取一行
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+50; int n,m,a[111][111]; int op[N], p[N], cnt; void add(int tp, int x) { ++cnt; op[cnt] = tp, p[cnt] = x; if (tp==1) REP(i,1,n) --a[i][x]; else REP(i,1,m) --a[x][i]; } int main() { scanf("%d%d",&n,&m); int sum = 0; REP(i,1,n) REP(j,1,m) scanf("%d",a[i]+j),sum+=a[i][j]; while (sum) { if (n>m) { int ok = 0; REP(i,1,m) { int s = 1e9; REP(j,1,n) s=min(s,a[j][i]); if (s) add(1,i),sum-=n,ok=1; } if (!ok) { REP(i,1,n) { int s = 1e9; REP(j,1,m) s=min(s,a[i][j]); if (s) add(2,i),sum-=m,ok=1; } if (!ok) return puts("-1"),0; } } else { int ok = 0; REP(i,1,n) { int s = 1e9; REP(j,1,m) s=min(s,a[i][j]); if (s) add(2,i),ok=1,sum-=m; } if (!ok) { REP(i,1,m) { int s = 1e9; REP(j,1,n) s=min(s,a[j][i]); if (s) add(1,i),ok=1,sum-=n; } if (!ok) return puts("-1"),0; } } } printf("%d ",cnt); REP(i,1,cnt) printf("%s %d ",op[i]==1?"col":"row",p[i]); }
2. 815B Karen and Test
大意: 有一个三角形的数表, 给定第一行, 下面每一行是上面相邻两数和或差, 求最底层的数是多少.
打表算一下每个数的贡献, 找下规律即可
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 200; vector<int> f[N][N]; int n, a[N]; vector<int> add(vector<int> a, vector<int> b) { REP(i,1,n) a[i]+=b[i]; return a; } vector<int> sub(vector<int> a, vector<int> b) { REP(i,1,n) a[i]-=b[i]; return a; } int main() { scanf("%d",&n); REP(i,1,n) REP(j,1,n) f[i][j].resize(n+1); REP(i,1,n) f[1][i][i] = 1; int cur = 1; REP(i,2,n) { REP(j,1,n-i+1) { if (cur) f[i][j]=add(f[i-1][j],f[i-1][j+1]); else f[i][j]=sub(f[i-1][j],f[i-1][j+1]); cur ^= 1; } } REP(i,1,n) printf("%d,",f[n][1][i]);hr; }
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+50; int n, a[N], fac[N], ifac[N]; int f[N], g[N]; ll C(int n, int m) { if (n<m) throw; return (ll)fac[n]*ifac[m]%P*ifac[n-m]%P; } int main() { fac[0]=1; REP(i,1,N-1) fac[i]=(ll)fac[i-1]*i%P; ifac[N-1]=inv(fac[N-1]); PER(i,0,N-2) ifac[i]=(ll)ifac[i+1]*(i+1)%P; scanf("%d", &n); REP(i,1,n) scanf("%d",a+i); if (n==1) return printf("%d ",a[1]),0; if (n%4==0) { int x = n/2-1; REP(i,1,n) g[i]=i&1?C(x,i/2):-C(x,i/2-1); } else if (n%4==1) { int x = (n-1)/2; REP(i,1,n) if (i&1) g[i] = C(x,i/2); } else if (n%4==2) { int x = (n-2)/2; REP(i,1,n) g[i]=C(x,(i-1)/2); } else { int x = (n-3)/2; REP(i,1,n-1) f[i]=C(x,(i-1)/2); REP(i,1,n) g[i]=i&1?f[i]-f[i-1]:f[i]+f[i-1]; } int ans = 0; REP(i,1,n) ans = (ans+(ll)g[i]*a[i])%P; if (ans<0) ans += P; printf("%d ", ans); }
3. 815C Karen and Supermarket
大意: $n$个商品构成一棵树, 商品原价$c_i$元, 用优惠券会减少$d_i$元, 但如果$i$使用优惠券, 那么必须购买$x_i$. 预算为$b$元, 求最多买多少商品.
暴力树形$dp$即可, 复杂度是$O(n^2)$, 因为每个二元组只会在lca出被枚举到一次
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 5e3+10; int n, b; int c[N],d[N],fa[N],sz[N]; int f[N][N], g[N][N]; vector<int> a[N]; //f[x][y] = x子树内, x用优惠券, 一共购买y个的最少花费 //g[x][y] = x子树内, x不用优惠券, 一共购买y个的最少花费 void chkmin(int &a, int b) {a>b?a=b:0;} void dfs(int x) { f[x][0] = g[x][0] = 0; for (int y:a[x]) { dfs(y); PER(i,0,sz[x]) REP(j,0,sz[y]) { chkmin(f[x][i+j],f[x][i]+min(g[y][j],f[y][j])); chkmin(g[x][i+j],g[x][i]+g[y][j]); } sz[x] += sz[y]; } ++sz[x]; PER(i,1,sz[x]) { g[x][i] = min(g[x][i],g[x][i-1]+c[x]); f[x][i] = f[x][i-1]+c[x]-d[x]; } } int main() { memset(f,0x3f,sizeof f); memset(g,0x3f,sizeof g); scanf("%d%d%d%d",&n,&b,c+1,d+1); REP(i,2,n) { scanf("%d%d%d",c+i,d+i,fa+i); a[fa[i]].pb(i); } dfs(1); PER(i,0,n) if (f[1][i]<=b||g[1][i]<=b) return printf("%d ",i),0; }
4. 815D Karen and Cards
大意: $n$张卡, 每张卡三个属性, 上限分别为$p,q,r$, 若一张卡存在两个属性值严格大于另一张卡, 那么这张卡能打败另一张卡. 求有多少张卡能打败其他所有卡.
按$c$排序, 从大到小枚举$c$, 对$a,b$建一个二维平面, 那么假设一张卡的$c_i<c$, 那么这张卡贡献就是对$1le xle a_i,1le yle b_i$的矩形赋零. 若$c_ige c$, 那么相当于对$1le xle p,1le yle b_i$和$1le xle a_i,1le yle q$的两个矩形赋零. 用线段树区间取max, 查询最大值即可.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head #ifdef ONLINE_JUDGE const int N = 1e6+50; #else const int N = 1e2+10; #endif int n, p, q, r; struct _ {int a,b,c;} f[N]; //区间取max, 区间求和 //维护区间min, 转化为区间赋值, 区间求和 struct { int ma,mi,tag; ll sum; void upd(int x, int t) { ma=mi=tag=x, sum=(ll)t*x; } } tr[N<<2]; void pu(int o) { tr[o].ma = max(tr[lc].ma,tr[rc].ma); tr[o].mi = min(tr[lc].mi,tr[rc].mi); tr[o].sum = tr[lc].sum+tr[rc].sum; tr[o].tag = 0; } void pd(int o, int l, int r) { if (tr[o].tag) { tr[lc].upd(tr[o].tag,mid-l+1); tr[rc].upd(tr[o].tag,r-mid); tr[o].tag = 0; } } void upd(int o, int l, int r, int ql, int qr, int v) { if (ql>qr||tr[o].mi>=v) return; if (ql<=l&&r<=qr&&tr[o].ma<=v) return tr[o].upd(v,r-l+1); pd(o,l,r); if (mid>=ql) upd(ls,ql,qr,v); if (mid<qr) upd(rs,ql,qr,v); pu(o); } int main() { scanf("%d%d%d%d", &n, &p, &q, &r); REP(i,1,n) scanf("%d%d%d",&f[i].a,&f[i].b,&f[i].c); REP(i,1,n) upd(1,1,p,1,f[i].a,f[i].b); sort(f+1,f+1+n,[](_ a,_ b){return a.c>b.c;}); ll ans = 0; int now = 1; PER(i,1,r) { while (now<=n&&f[now].c>=i) { upd(1,1,p,1,f[now].a,q); upd(1,1,p,f[now].a+1,p,f[now].b); ++now; } ans += (ll)p*q-tr[1].sum; } printf("%lld ", ans); }
5. 815E Karen and Neighborhood
大意: $n$个房间,$k$个人轮流去住, 第一个人住$1$号, 之后每个人会住距离有人的房间最远的房间, 有多个的话住编号最小的, 求第$k$个人房间号.