2014浙大校赛
ZOJ 3767 Elevator
求和，签到题
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; #define pb(a) push(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c)); } template<class T> T max(const T& a,const T& b,const T& c) { return max(max(a,b),max(a,c)); } void debug() { #ifdef ONLINE_JUDGE #else freopen("in.txt","r",stdin); //freopen("d:\\out1.txt","w",stdout); #endif } int getch() { int ch; while((ch=getchar())!=EOF) { if(ch!=' '&&ch!='\n')return ch; } return EOF; } int main() { int n,m; int t; scanf("%d",&t); for(int ca=1;ca<=t;ca++) { scanf("%d%d",&n,&m); int res=0; for(int i=0;i<n;i++) { int x;scanf("%d",&x); res+=x; } printf("%s\n",res<=m?"Safe":"Warning"); } return 0; }
ZOJ 3775 ?(>_o)!
题意中一大堆无用信息，其实就是判断源码和输出是否一样。简单字符串处理，string很方便
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; #define pb(a) push(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c)); } template<class T> T max(const T& a,const T& b,const T& c) { return max(max(a,b),max(a,c)); } void debug() { #ifdef ONLINE_JUDGE #else freopen("in.txt","r",stdin); //freopen("d:\\out1.txt","w",stdout); #endif } int getch() { int ch; while((ch=getchar())!=EOF) { if(ch!=' '&&ch!='\n')return ch; } return EOF; } int main() { int t; cin>>t; getchar(); for(int ca=1;ca<=t;ca++) { string s1; getline(cin,s1); string s2; for(int i=0;i<s1.length();i++) { if(s1[i]=='_') s2+=s1; if(s1[i]=='!') s2+="Hello, world!"; } // cout<<s1<<endl<<s2<<endl; cout<<(s1==s2?"Yes":"No")<<endl; } return 0; }
ZOJ 3770 Ranking System
按比例分等级，排序
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; #define pb(a) push(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c)); } template<class T> T max(const T& a,const T& b,const T& c) { return max(max(a,b),max(a,c)); } void debug() { #ifdef ONLINE_JUDGE #else freopen("in.txt","r",stdin); //freopen("d:\\out1.txt","w",stdout); #endif } int getch() { int ch; while((ch=getchar())!=EOF) { if(ch!=' '&&ch!='\n')return ch; } return EOF; } struct number { int id; string date; int score; int order; int lv; bool operator < (const number &ant) const { if(score!=ant.score)return score>ant.score; else if(date!=ant.date) return date<ant.date; else return id<ant.id; } }da[2005]; bool cmp(const number &a,const number &b) { return a.order<b.order; } int main() { int t; cin>>t; for(int ca=1;ca<=t;ca++) { int n;cin>>n; for(int i=0;i<n;i++) { cin>>da[i].id>>da[i].date>>da[i].score; da[i].order=i; } sort(da,da+n); int num=0; while(num<n&&da[num].score>0)num++; int pre=0; int x=(int)(num*0.03); for(int i=pre;i<pre+x;i++) da[i].lv=6; pre=pre+x; x=(int)(num*0.07); for(int i=pre;i<pre+x;i++) da[i].lv=5; pre=pre+x; x=(int)(num*0.2); for(int i=pre;i<pre+x;i++) da[i].lv=4; pre=pre+x; x=(int)(num*0.3); for(int i=pre;i<pre+x;i++) da[i].lv=3; pre=pre+x; for(;pre<n&&da[pre].score>0;pre++) da[pre].lv=2; for(;pre<n;pre++) da[pre].lv=1; sort(da,da+n,cmp); for(int i=0;i<n;i++) printf("LV%d\n",da[i].lv); } return 0; }
ZOJ 3768 Continuous Login
将n分解为尽量少个前缀和，搜索，暴力出奇迹，我用的是迭代加深
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; #define pb(a) push(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c)); } template<class T> T max(const T& a,const T& b,const T& c) { return max(max(a,b),max(a,c)); } void debug() { #ifdef ONLINE_JUDGE #else freopen("in.txt","r",stdin); //freopen("d:\\out1.txt","w",stdout); #endif } int getch() { int ch; while((ch=getchar())!=EOF) { if(ch!=' '&&ch!='\n')return ch; } return EOF; } const int maxn=40000; int f[maxn]; int ans[maxn]; int cal(int n) { int k=(int)(sqrt(n*2)+1); while(f[k]>n)k--; return k; } int dfs(int d,int n,int k) { if(k==1) { ans[k]=cal(n); if(n==f[cal(n)])return true; }else { for(int x=min(d,cal(n));x>=1;x--) { ans[k]=x; if(dfs(x,n-f[x],k-1))return true; } } return 0; } int main() { for(int i=1;i<maxn;i++) f[i]=i*(i+1)/2; int t; scanf("%d",&t); for(int ca=1;ca<=t;ca++) { int n; scanf("%d",&n); int res; for(int i=1;;i++) { if(dfs(cal(n),n,i)) { res=i;break; } } for(int i=1;i<=res;i++) printf("%d%c",ans[i],i==res?'\n':' '); } return 0; }
ZOJ 3772 Calculate the Function
做法1：线段树维护区间矩阵乘积，注意是左乘还是右乘
做法2：预处理出全部前缀矩阵乘积和全部前缀逆矩阵乘积（就跟区间和sum(b)-sum(a-1)一样）
线段树：
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; #define pb(a) push(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c)); } template<class T> T max(const T& a,const T& b,const T& c) { return max(max(a,b),max(a,c)); } void debug() { #ifdef ONLINE_JUDGE #else freopen("in.txt","r",stdin); //freopen("d:\\out1.txt","w",stdout); #endif } int getch() { int ch; while((ch=getchar())!=EOF) { if(ch!=' '&&ch!='\n')return ch; } return EOF; } const int mod=1000000007; struct Matrix { ll a[2][2]; Matrix(){memset(a,0,sizeof(a));} void set(ll A) { a[0][0]=a[1][0]=1; a[0][1]=A; a[1][1]=0; } void setI() { a[0][0]=a[1][1]=1; a[0][1]=a[1][0]=0; } }; Matrix operator * (const Matrix &a,const Matrix &b) { Matrix res; for(int i=0;i<2;i++) for(int j=0;j<2;j++) for(int k=0;k<2;k++) res.a[i][j]=(res.a[i][j]+a.a[i][k]*b.a[k][j])%mod; return res; } const int maxn=100005; ll A[maxn]; Matrix v[maxn<<2]; void PushUp(int idx) { v[idx]=v[idx<<1|1]*v[idx<<1]; } void build(int idx,int l,int r) { if(l==r) { v[idx].set(A[r]); }else { int mid=(r+l)>>1; build(lson); build(rson); PushUp(idx); } } Matrix query(int idx,int l,int r,int tl,int tr) { if(tl<=l&&tr>=r) return v[idx]; Matrix res; res.setI(); int mid=(r+l)>>1; if(tl<=mid) res=res*query(lson,tl,tr); if(tr>mid) res=query(rson,tl,tr)*res; return res; } ll cal(const Matrix &x,int a,int b) { return (x.a[0][0]*a+x.a[0][1]*b)%mod; } int main() { int t; scanf("%d",&t); for(int ca=1;ca<=t;ca++) { int n,q; scanf("%d%d",&n,&q); for(int i=1;i<=n;i++) scanf("%lld",&A[i]); build(1,1,n); for(int Q=1;Q<=q;Q++) { int l,r; scanf("%d%d",&l,&r); ll res=0; if(r-l<2) res=A[r]; else { Matrix x=query(1,1,n,l+2,r); res=cal(x,A[l+1],A[l]); } printf("%lld\n",res); } } return 0; }
ZOJ 3769 Diablo III
背包，先把特殊的装备处理了，把任意一对Finger当成一件装备还有把Weapon 和Shield 的任意组合当成Two-Handed
然后就是普通的背包了，dp[i][j]表示装备前i类装备防御力为j的最大攻击力(j>=m的情况都当成是m)
需要优化：把攻击力和防御力都低的舍掉。
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; #define pb(a) push(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c)); } template<class T> T max(const T& a,const T& b,const T& c) { return max(max(a,b),max(a,c)); } void debug() { #ifdef ONLINE_JUDGE #else freopen("in.txt","r",stdin); //freopen("d:\\out1.txt","w",stdout); #endif } int getch() { int ch; while((ch=getchar())!=EOF) { if(ch!=' '&&ch!='\n')return ch; } return EOF; } struct attri { int a,b; }; map<string,int> id; int ecnt; int ID(string s) { if(id.find(s)==id.end()) id[s]=++ecnt; return id[s]; } vector<attri> da[15]; void deal() { //combine weapon and shield int a=ID(string("Weapon")); da[a].push_back((attri){0,0}); int b=ID(string("Shield")); da[b].push_back((attri){0,0}); int c=ID(string("Two-Handed")); for(int i=0;i<da[a].size();i++) for(int j=0;j<da[b].size();j++) da[c].push_back((attri){ da[a][i].a+da[b][j].a, da[a][i].b+da[b][j].b }); da[a].clear(); da[b].clear(); //combine two finger int d=ID(string("Finger")); da[d].push_back((attri){0,0}); vector<attri> buf; for(int i=0;i<da[d].size();i++) for(int j=i+1;j<da[d].size();j++) buf.push_back((attri){ da[d][i].a+da[d][j].a, da[d][i].b+da[d][j].b}); da[d]=buf; for(int i=1;i<=13;i++) da[i].push_back((attri){0,0}); } int dp[15][50005]; int main() { std::ios::sync_with_stdio(false); int t; cin>>t; for(int ca=1;ca<=t;ca++) { int n,m; cin>>n>>m; ecnt=0; id.clear(); for(int i=1;i<=13;i++) da[i].clear(); for(int i=1;i<=n;i++) { string s; int a,b; cin>>s>>a>>b; int id=ID(s); da[id].push_back((attri){a,b}); } deal(); memset(dp,-1,sizeof(dp)); dp[0][0]=0; // for(int i=1;i<=13;i++) // { // cout<<"ID:"<<i<<endl; // for(int j=0;j<da[i].size();j++) // cout<<da[i][j].a<<' '<<da[i][j].b<<endl; // // } for(int i=1;i<=13;i++) { int maxa=0,maxb=0; for(int j=0;j<da[i].size();j++) { if(da[i][j].a<maxa&&da[i][j].b<maxb)continue; if(da[i][j].a>maxa&&da[i][j].b>maxb) { maxa=da[i][j].a;maxb=da[i][j].b; } for(int d=0;d<=m;d++)if(dp[i-1][d]!=-1) { int b=d+da[i][j].b; if(b>=m)b=m; dp[i][b]=max(dp[i][b],dp[i-1][d]+da[i][j].a); } } } cout<<dp[13][m]<<endl; } return 0; }
目前会做的只有这6题了。。
