nenu contest

http://vjudge.net/vjudge/contest/view.action?cid=54393#overview

A n^2能过 对第二个n我二分了一下，快了一点点，nlogn

 #include<cstdio>
 #include<algorithm>
 using namespace std;
 double judge[];
 int main(){
     for(int i=;i<;i++){
         judge[i]=i*;
     }
     double now;
     int n,s;
     while(~scanf("%d%d",&n,&s)){
         int sum=;
         for(int i=;i<n;i++){
             scanf("%lf",&now);
             int L=lower_bound(judge,judge+,now)-judge;
             if(judge[L]>now) L--;
             if(judge[L]<=now&&now<=judge[L]+) sum++;
         }
         printf("%.1f\n",sum*(+0.2*s));
     }
     return ;
 }

A LEE的on的方法也很值得借鉴，竟然比我二分的慢，不知为何，数据小吧

 #include<cstdio>
 const double eps = 1e-;
 int main() {
     int n, s, cnt;
     double t;
     while (~scanf("%d%d", &n, &s)) {
         cnt = ;
         while (n--) {
             scanf("%lf", &t);
             int T = int(t);
             if (T %  <  || (T %  ==  && (t-T) < eps)) {
                 cnt++;
             }
         }
         printf("%.1f\n", cnt*( + 0.2*s));
     }
     return ;
 }

B %X scanf printf 16进制输入输出

 #include<cstdio>
 #include<cstring>
 const int M=;
 char op[M];
 int main(){
     int t,sa,sb;
     while(~scanf("%d",&t)){
         while(t--){
             scanf("%X%X",&sa,&sb);
             getchar();
             gets(op);
             int lp=strlen(op);
             for(int i=;i<lp;i++){
                 if(op[i]==''){
                     if(op[i+]=='A'){
                         sa+=sb;
                     }
                     else{
                         sb+=sa;
                     }
                     i+=;
                 }
                 else if(op[i]=='F'){
                     if(op[i+]=='A'){
                         sa+=sa;
                     }
                     else{
                         sb+=sb;
                     }
                     i+=;
                 }
             }
             printf("%X %X\n",sa,sb);
         }
     }
     return ;
 }

C 比赛时n^2logn 900ms压线过，事实上nlogn就行了

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<map>
 #define mt(a,b) memset(a,b,sizeof(a))
 using namespace std;
 const int M=;
 class UnionFindSet {  //并查集
     int par[M];
 public:
     void init() {
         mt(par,-);
     }
     int getroot(int x) {
         int i=x,j=x,temp;
         while(par[i]>=) i=par[i];
         while(j!=i) {
             temp=par[j];
             par[j]=i;
             j=temp;
         }
         return i;
     }
     bool unite(int x,int y) {
         int p=getroot(x);
         int q=getroot(y);
         if(p==q)return false;
         if(par[p]>par[q]) {
             par[q]+=par[p];
             par[p]=q;
         } else {
             par[p]+=par[q];
             par[q]=p;
         }
         return true;
     }
 }gx;
 map<string,int> mp;
 string str;
 int main(){
     int n,m;
     char op[];
     while(~scanf("%d",&n)){
         bool flag=false;
         gx.init();
         mp.clear();
         for(int i=;i<=n;i++){
             scanf("%d",&m);
             while(m--){
                 flag=true;
                 scanf("%s",op);
                 str=(string)op;
                 if(!mp[str]){
                     mp[str]=i;
                 }
                 else{
                     gx.unite(mp[str],i);
                 }
             }
         }
         if(!flag){
             printf("%d\n",n);
             continue;
         }
         int ans=;
         for(int i=;i<=n;i++){
             if(i==gx.getroot(i)) ans++;
         }
         printf("%d\n",ans-);
     }
     return ;
 }

I n^2能过400ms on单调队列62ms

 #include<cstdio>
 #include<cstdlib>
 #include<algorithm>
 using namespace std;
 const int M=;
 int a[M],b[M];
 int main(){
     int n,m;
     while(~scanf("%d%d",&n,&m)){
         for(int i=;i<n;i++){
             scanf("%d",&a[i]);
         }
         for(int i=;i<m;i++){
             scanf("%d",&b[i]);
         }
         int la=,lb=,ans=1e9;
         while(la!=n&&lb!=m){
             ans=min(ans,abs(a[la]-b[lb]));
             if(a[la]>b[lb]) lb++;
             else la++;
         }
         printf("%d\n",ans);
     }
     return ;
 }

J  on 水题

 #include<cstdio>
 typedef __int64 LL;
 const int M=;
 char a[M][];
 int main(){
     int n;
     while(~scanf("%d",&n)){
         for(int i=;i<n;i++){
             scanf("%s",a[i]);
         }
         LL ans=;
         LL sum=;
         for(int i=n-;i>=;i--){
             if(a[i][]=='W') ans+=sum;
             else sum++;
         }
         printf("%I64d\n",ans);
     }
     return ;
 }