Codeforces Round #250 (Div. 2) A B C

C 贪心 写的时候突然发现这么容易，所有的绳子都要拆掉，而且绳子的个数固定，所以只要每次拆绳子，只要找绳子两端v小的即可，O(n)  //代码里面有没用的冗余

//#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <iostream>#include <iomanip>#include <cmath>#include <map>#include <set>#include <queue>using namespace std;#define ls(rt) rt*2#define rs(rt) rt*2+1#define ll long long#define ull unsigned long long#define rep(i,s,e) for(int i=s;i<e;i++)#define repe(i,s,e) for(int i=s;i<=e;i++)#define CL(a,b) memset(a,b,sizeof(a))#define IN(s) freopen(s,"r",stdin)#define OUT(s) freopen(s,"w",stdout)const ll ll_INF = ((ull)(-1))>>1;const double EPS = 1e-8;const double pi = acos(-1.0);const int INF = 100000000;const int MAXN = 2000*2+100;vector<int>g[MAXN];int v[MAXN],id[MAXN];int n,m;int main(){    //IN("C.txt");    while(~scanf("%d%d",&n,&m))    {        int u,t;        for(int i=0;i<=n;i++)            g[i].clear();        for(int i=1;i<=n;i++)            scanf("%d",&v[i]),id[i]=i;        //sort(id+1,id)        ll ans=0;        for(int i=0;i<m;i++)        {            scanf("%d%d",&u,&t);            g[u].push_back(t);            g[t].push_back(u);            if(v[t]>v[u])ans+=v[u];            else ans+=v[t];        }        printf("%I64d\n",ans);    }    return 0;}

B---胡蒙的，至今不解为啥按lowbit从大到小一定可以找出sum的组合

//#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <iostream>#include <iomanip>#include <cmath>#include <map>#include <set>#include <queue>using namespace std;#define ls(rt) rt*2#define rs(rt) rt*2+1#define ll long long#define ull unsigned long long#define rep(i,s,e) for(int i=s;i<e;i++)#define repe(i,s,e) for(int i=s;i<=e;i++)#define CL(a,b) memset(a,b,sizeof(a))#define IN(s) freopen(s,"r",stdin)#define OUT(s) freopen(s,"w",stdout)const ll ll_INF = ((ull)(-1))>>1;const double EPS = 1e-8;const double pi = acos(-1.0);const int INF = 100000000;const int MAXN = 1e5+100;int vis[MAXN],a[MAXN];int sum,up;inline int lowbit(int x){    return x&(-x);}vector<int>ans;bool cmp(int ca,int cb){    return lowbit(ca)>lowbit(cb);}int main(){    //IN("B.txt");    while(~scanf("%d%d",&sum,&up))    {        ans.clear();        for(int i=0;i<=up;i++)            a[i]=i;        sort(a+1,a+1+up,cmp);        for(int i=1;i<=up;i++)         {             int tmp=lowbit(a[i]);            // printf("i=%d %d\n",i,a[i]);             /////             if(sum>=tmp)sum-=tmp,ans.push_back(a[i]);             if(sum==0)break;         }         if(sum!=0)puts("-1");         else         {             printf("%d\n",ans.size());             if(ans.size())printf("%d",ans[0]);             for(int i=1;i<ans.size();i++)                printf(" %d",ans[i]);             putchar('\n');         }    }    return 0;}

A  纯属联系string类的substr函数了 不过用String数组写更好些，我的代码冗余严重

//#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <iostream>#include <iomanip>#include <cmath>#include <map>#include <set>#include <queue>using namespace std;#define ls(rt) rt*2#define rs(rt) rt*2+1#define ll long long#define ull unsigned long long#define rep(i,s,e) for(int i=s;i<e;i++)#define repe(i,s,e) for(int i=s;i<=e;i++)#define CL(a,b) memset(a,b,sizeof(a))#define IN(s) freopen(s,"r",stdin)#define OUT(s) freopen(s,"w",stdout)const ll ll_INF = ((ull)(-1))>>1;const double EPS = 1e-8;const double pi = acos(-1.0);const int INF = 100000000;string a,b,c,d;int checka(){    int len=a.size()*2;    if(len <= b.size() && len <=c.size() && len<=d.size())return 1;    if(a.size()>=b.size()*2 && a.size()>=d.size()*2 && a.size()>=c.size()*2)return 1;    return 0;}int checkb(){    int len=b.size()*2;    if(len <= a.size() && len <=c.size() && len<=d.size())return 1;    if(b.size()>=a.size()*2 && b.size()>=d.size()*2 && b.size()>=c.size()*2)return 1;    return 0;}int checkc(){    int len=c.size()*2;    if(len <= a.size() && len <=c.size() && len<=d.size())return 1;    if(c.size()>=a.size()*2 && c.size()>=d.size()*2 && c.size()>=b.size()*2)return 1;    return 0;}int checkd(){    int len=d.size()*2;    if(len <= a.size() && len <=c.size() && len<=b.size())return 1;    if(d.size()>=a.size()*2 && d.size()>=c.size()*2 && d.size()>=b.size()*2)return 1;    return 0;}int main(){    //IN("A.txt");    while(cin >> a >> b >> c >> d)    {        a=a.substr(2,a.size()-2);        b=b.substr(2,b.size()-2);        c=c.substr(2,c.size()-2);        d=d.substr(2,d.size()-2);        int flag=0;        char ans;        if(checka()){ans='A';flag++;}        if(checkb()){ans='B';flag++;}        if(checkc()){ans='C';flag++;}        if(checkd()){ans='D';flag++;}        if(flag == 1){printf("%c\n",ans);continue;}        puts("C");    }    return 0;}