7月1日 cf总结
7月1日 cf总结
A题:水题,题意看半天,30分钟才A。
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<set> #include<map> #include<string> #include<math.h> #include<cctype> #define ll long long #define REP(i,a,b) for(int (i)=(a);(i)<=(b);(i)++) #define REPP(i,a,b,t) for(int (i)=(a);(i)<=(b);(i)+=(t)) #define rep(i,a,b) for(int (i)=(a);(i)>=(b);(i)--) #define repp(i,a,b,t) for(int (i)=(a);(i)>=(b);(i)-=(t)) #define PII pair<int,int> #define fst first #define snd second #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&(x)) #define RII(x,y) scanf("%d%d",&(x),&(y)) #define RIII(x,y,z) scanf("%d%d%d",&(x),&(y),&(z)) #define DRI(x) int (x);scanf("%d",&(x)) #define DRII(x,y) int (x),(y);scanf("%d%d",&(x),&(y)) #define DRIII(x,y,z) int (x),(y),(z);scanf("%d%d",&(x),&(y),&(z)) #define RS(x) scanf("%s",x) #define RSS(x,y) scanf("%s%s",x,y) #define DRS(x) char x[maxn];scanf("%s",x) #define DRSS(x,y) char x[maxn],y[maxn];scanf("%s%s",x,y) #define MS0(a) memset((a),0,sizeof((a))) #define MS1(a) memset((a),-1,sizeof((a))) #define MS(a,b) memset((a),(b),sizeof((a))) #define ALL(v) v.begin(),v.end() #define SZ(v) (int)(v).size() using namespace std; const int maxn=1000100; const int INF=(1<<29); const double EPS=0.0000000001; const double Pi=acos(-1.0); int a[10],b[10]; int n; int c[10]; int main() { while(cin>>n){ REP(i,1,3) RI(a[i]),RI(b[i]); int ans=a[1]+a[2]+a[3]; REP(i,1,3) c[i]=a[i]; REP(i,1,3){ if(ans==n) break; REP(j,1,b[i]-a[i]){ ans++; c[i]++; if(ans==n) break; } } cout<<c[1]<<" "<<c[2]<<" "<<c[3]<<endl; } return 0; }
B题:水题,题意还是看半天,坑的是cout输出浮点数在第32个样例WA了,该printf就过了。。。。。看来以后浮点数都用printf输出。。。
由于cout对浮点数只保留到小数点后6位,而printf可以保留的位数更多,且可以根据需要确定进度,看来以后浮点数直接printf就好了。。
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<set> #include<map> #include<string> #include<math.h> #include<cctype> #define ll long long #define REP(i,a,b) for(int (i)=(a);(i)<=(b);(i)++) #define REPP(i,a,b,t) for(int (i)=(a);(i)<=(b);(i)+=(t)) #define rep(i,a,b) for(int (i)=(a);(i)>=(b);(i)--) #define repp(i,a,b,t) for(int (i)=(a);(i)>=(b);(i)-=(t)) #define PII pair<int,int> #define fst first #define snd second #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&(x)) #define RII(x,y) scanf("%d%d",&(x),&(y)) #define RIII(x,y,z) scanf("%d%d%d",&(x),&(y),&(z)) #define DRI(x) int (x);scanf("%d",&(x)) #define DRII(x,y) int (x),(y);scanf("%d%d",&(x),&(y)) #define DRIII(x,y,z) int (x),(y),(z);scanf("%d%d",&(x),&(y),&(z)) #define RS(x) scanf("%s",x) #define RSS(x,y) scanf("%s%s",x,y) #define DRS(x) char x[maxn];scanf("%s",x) #define DRSS(x,y) char x[maxn],y[maxn];scanf("%s%s",x,y) #define MS0(a) memset((a),0,sizeof((a))) #define MS1(a) memset((a),-1,sizeof((a))) #define MS(a,b) memset((a),(b),sizeof((a))) #define ALL(v) v.begin(),v.end() #define SZ(v) (int)(v).size() using namespace std; const int maxn=1000100; const int INF=(1<<29); const double EPS=0.0000000001; const double Pi=acos(-1.0); double c[maxn]; int n; double w; int main() { while(cin>>n>>w){ REP(i,1,2*n) scanf("%lf",&c[i]); sort(c+1,c+1+2*n); double a=c[1],b=c[n+1]; double ans=0; if(a*2<b){ ans=a*n+a*2*n; if(ans<w) printf("%f\n",ans); else printf("%f\n",w); } else{ ans=b*n/2.0+b*n; if(ans<w) printf("%f\n",ans); else printf("%f\n",w); } } return 0; }
没有AC不了的题,只有不努力的ACMER!