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[128];
int main(){
    for(int i=0;i<100;i++){
        judge[i]=i*10;
    }
    double now;
    int n,s;
    while(~scanf("%d%d",&n,&s)){
        int sum=0;
        for(int i=0;i<n;i++){
            scanf("%lf",&now);
            int L=lower_bound(judge,judge+100,now)-judge;
            if(judge[L]>now) L--;
            if(judge[L]<=now&&now<=judge[L]+5) sum++;
        }
        printf("%.1f
",sum*(60+0.2*s));
    }
    return 0;
}

 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-8;
int main() {
    int n, s, cnt;
    double t;
    while (~scanf("%d%d", &n, &s)) {
        cnt = 0;
        while (n--) {
            scanf("%lf", &t);
            int T = int(t);
            if (T % 10 < 5 || (T % 10 == 5 && (t-T) < eps)) {
                cnt++;
            }
        }
        printf("%.1f
", cnt*(60 + 0.2*s));
    }
    return 0;
}

B %X scanf printf 16进制输入输出

#include<cstdio>
#include<cstring>
const int M=1010;
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=0;i<lp;i++){
                if(op[i]=='8'){
                    if(op[i+3]=='A'){
                        sa+=sb;
                    }
                    else{
                        sb+=sa;
                    }
                    i+=8;
                }
                else if(op[i]=='F'){
                    if(op[i+3]=='A'){
                        sa+=sa;
                    }
                    else{
                        sb+=sb;
                    }
                    i+=5;
                }
            }
            printf("%X %X
",sa,sb);
        }
    }
    return 0;
}

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=128;
class UnionFindSet {  //并查集
    int par[M];
public:
    void init() {
        mt(par,-1);
    }
    int getroot(int x) {
        int i=x,j=x,temp;
        while(par[i]>=0) 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[32];
    while(~scanf("%d",&n)){
        bool flag=false;
        gx.init();
        mp.clear();
        for(int i=1;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);
            continue;
        }
        int ans=0;
        for(int i=1;i<=n;i++){
            if(i==gx.getroot(i)) ans++;
        }
        printf("%d
",ans-1);
    }
    return 0;
}

I n^2能过400ms on单调队列62ms

#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int M=1010;
int a[M],b[M];
int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
        }
        for(int i=0;i<m;i++){
            scanf("%d",&b[i]);
        }
        int la=0,lb=0,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
",ans);
    }
    return 0;
}

J  on 水题

#include<cstdio>
typedef __int64 LL;
const int M=1000010;
char a[M][4];
int main(){
    int n;
    while(~scanf("%d",&n)){
        for(int i=0;i<n;i++){
            scanf("%s",a[i]);
        }
        LL ans=0;
        LL sum=0;
        for(int i=n-1;i>=0;i--){
            if(a[i][0]=='W') ans+=sum;
            else sum++;
        }
        printf("%I64d
",ans);
    }
    return 0;
}