A
Problem description
给出数a,b,n,在a后加上1个数,并使加后的数是b的倍数.输出操作n次后的数.
Data Limit：1 ≤ a, b, n ≤ 105 Time Limit: 1s
Solution
易得只要第一次成功,剩下加0,就可以了
Code
#include<cstdio> long long a,b,n,i,j; long long ans[500000];bool bo=false; int main() { scanf("%lld%lld%lld",&a,&b,&n);//a=a%b; for (i=0;i<=9;i++) { if ((a*10+i)%b==0) { a=a*10+i;bo=true; break; } } if (bo) { printf("%lld",a); for (i=1;i<=n-1;i++)printf("%d",0); }else printf("%d",-1); return 0; }
B
Problem description
输入=一个字符串,输出出现最多的合法日期.
Data Limit：|s|<=10e5 Time Limit: 1s
Solution
暴力枚举每10个连续字符,判断是否是合法日期并计数即可.
Code
var s,t:ansistring; n,i,j,d,m,y,k,max:longint; a:array[1..31,1..12,2013..2015]of longint; day:array[1..12] of longint; begin readln(s); day[1]:=31; day[2]:=28; day[3]:=31; day[4]:=30; day[5]:=31; day[6]:=30; day[7]:=31; day[8]:=31; day[9]:=30; day[10]:=31; day[11]:=30; day[12]:=31; max:=-10; for i:=1 to length(s) do if (s[i]='-')and(s[i+3]='-')and((s[i-2]<='9')and(s[i-2]>='0')) and((s[i-1]<='9')and(s[i-1]>='0'))and((s[i+1]<='9')and(s[i+1]>='0')) and((s[i+2]<='9')and(s[i+2]>='0'))and((s[i+4]<='9')and(s[i+4]>='0')) and((s[i+5]<='9')and(s[i+5]>='0'))and((s[i+6]<='9')and(s[i+6]>='0')) and((s[i+7]<='9')and(s[i+7]>='0'))then begin d:=(ord(s[i-2])-48)*10+ord(s[i-1])-48; m:=(ord(s[i+1])-48)*10+ord(s[i+2])-48; y:=(ord(s[i+4])-48)*1000+(ord(s[i+5])-48)*100+(ord(s[i+6])-48)*10+ord(s[i+7])-48; if (y<=2015)and(y>=2013)and(m<=12)and(m>=1) then if (d>=1)and(d<=day[m]) then begin inc(a[d,m,y]); end; end; for i:=1 to 31 do for j:=1 to 12 do for k:=2013 to 2015 do if a[i,j,k]>max then begin max:=a[i,j,k]; d:=i;m:=j;y:=k; end; if d<10 then write('0',d) else write(d); write('-'); if m<10 then write('0',m) else write(m); write('-'); writeln(y); end.
C
Problem description
有n个箱子,进行一次操作,把一个箱子的球拿出来,然后放到后面的箱子里,每个箱子放一个,直到取的球放完. 若放到了第n个箱子,下一个放第1个箱子.给出操作后的情况和最后放球的箱子,求出原来的情况.
Data Limit：2 ≤ n ≤ 105 Time Limit: 1s
Solution
现在球数最少的的箱子,这就是原来选的箱子,然后就可以求出原来的样子啦.
Code
#include<cstdio> long long n,i,j,x,min=1e18,ans=0; long long a[500000]; int main() { scanf("%lld%lld",&n,&x); for (i=1;i<=n;i++) { scanf("%lld",&a[i]); if (a[i]<min)min=a[i]; } if (min>0)min--; for (i=1;i<=n;i++)a[i]=a[i]-min; while (a[x]>0) { ans++;a[x]--;x--; if (x==0)x=n; } a[x]=min*n+ans; for (i=1;i<=n;i++)printf("%lld ",a[i]); return 0; }
D
Problem description
给出n个点的颜色(黑或白)和权值(所有与其相连的边的权值和)求出原来的树.
Data Limit：n <= 1e5 Time Limit: 1s
Solution
贪心,每次枚举两个点,连权值最大的边即可.
Code
#include<cstdio> #include<algorithm> using namespace std; long long n,i,j,t,topb,topw,ans[2000000][4],topa; bool bo=false; struct point { long long a,sum; }; bool bbb(point x,point y) { return x.sum>y.sum; } point black[2000000],white[2000000]; int main() { scanf("%lld",&n); for (i=1;i<=n;i++) { scanf("%lld",&t); if (t==0) { topw++; scanf("%lld",&white[topw].sum); white[topw].a=i; }else { topb++; scanf("%lld",&black[topb].sum); black[topb].a=i; } } i=1;j=1;topa=0; while ((i<=topb)&&(j<=topw)) { topa++; ans[topa][1]=black[i].a; ans[topa][2]=white[j].a; t=min(white[j].sum,black[i].sum); ans[topa][3]=t; white[j].sum-=t;black[i].sum-=t; if (black[i].sum) j++; else if(white[j].sum) i++; else if(i<topb) i++; else j++; } for (i=1;i<=n-1;i++) printf("%lld %lld %lld ",ans[i][1],ans[i][2],ans[i][3]); return 0; }
E
Problem description
给出一个平面上的n个点,用两条平行于x轴的直线和两条平行于y轴的直线把平面分成9块 使每个块中的点符合给出的要求.
Data Limit：9 ≤ n ≤ 10^5 Time Limit:2s
Solution
先全排列出每个块的点数然后用线段树求出能否满足题意
Code
#include<iostream> #include<cstdio> #include<map> #include<cstring> #include<cmath> #include<vector> #include<algorithm> #include<set> #include<string> #include<queue> #define inf 1000000005 #define M 40 #define N 100005 #define maxn 300005 #define eps 1e-12 #define zero(a) fabs(a)<eps #define Min(a,b) ((a)<(b)?(a):(b)) #define Max(a,b) ((a)>(b)?(a):(b)) #define pb(a) push_back(a) #define mp(a,b) make_pair(a,b) #define mem(a,b) memset(a,b,sizeof(a)) #define LL long long #define MOD 1000000007 #define lson step<<1 #define rson step<<1|1 #define sqr(a) ((a)*(a)) #define Key_value ch[ch[root][1]][0] #define test puts("OK"); #define pi acos(-1.0) #define lowbit(x) ((-(x))&(x)) #define HASH1 1331 #define HASH2 10001 #pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; struct Set_tree{ int left,right; vector<int>v; }L[N*4]; struct Point{ int x,y; bool operator<(const Point n)const{ return x!=n.x?x<n.x:y<n.y; } }p[N]; int n,x[N],y[N]; int a[9],b[9]; double ret_x1,ret_x2,ret_y1,ret_y2; void Bulid(int step,int l,int r){ L[step].left=l; L[step].right=r; L[step].v.clear(); for(int i=l;i<=r;i++) L[step].v.pb(p[i].y); sort(L[step].v.begin(),L[step].v.end()); if(l==r) return; int m=(l+r)>>1; Bulid(lson,l,m); Bulid(rson,m+1,r); } int Query(int step,int l,int r,int val){ if(L[step].left==l&&r==L[step].right){ if(L[step].v.size()==0) return 0; if(L[step].v[0]>val) return 0; if(L[step].v.back()<=val) return L[step].v.size(); return (upper_bound(L[step].v.begin(),L[step].v.end(),val)-L[step].v.begin()); } int m=(L[step].left+L[step].right)>>1; if(r<=m) return Query(lson,l,r,val); else if(l>m) return Query(rson,l,r,val); else return Query(lson,l,m,val)+Query(rson,m+1,r,val); } bool ok(){ int x1=b[a[0]]+b[a[1]]+b[a[2]]-1; int x2=x1+b[a[3]]+b[a[4]]+b[a[5]]; int y1=b[a[0]]+b[a[3]]+b[a[6]]-1; int y2=y1+b[a[1]]+b[a[4]]+b[a[7]]; if(x1+1>=n||x[x1]==x[x1+1]) return false; if(x2+1>=n||x[x2]==x[x2+1]) return false; if(y1+1>=n||y[y1]==y[y1+1]) return false; if(y2+1>=n||y[y2]==y[y2+1]) return false; if(Query(1,0,x1,y[y1])!=b[a[0]]) return false; if(Query(1,0,x1,y[y2])!=b[a[0]]+b[a[1]]) return false; if(Query(1,x1+1,x2,y[y1])!=b[a[3]]) return false; if(Query(1,x1+1,x2,y[y2])!=b[a[3]]+b[a[4]]) return false; ret_x1=(x[x1]+x[x1+1])/2.0; ret_x2=(x[x2]+x[x2+1])/2.0; ret_y1=(y[y1]+y[y1+1])/2.0; ret_y2=(y[y2]+y[y2+1])/2.0; return true; } int main(){ //freopen("input.txt","r",stdin); while(scanf("%d",&n)!=EOF){ for(int i=0;i<n;i++){ scanf("%d%d",&p[i].x,&p[i].y); x[i]=p[i].x;y[i]=p[i].y; } sort(p,p+n); sort(x,x+n); sort(y,y+n); Bulid(1,0,n-1); for(int i=0;i<9;i++) scanf("%d",&b[i]); for(int i=0;i<9;i++) a[i]=i; int t=362880; while(t--){ if(ok()){ printf("%.1f %.1f %.1f %.1f ",ret_x1,ret_x2,ret_y1,ret_y2); break; } next_permutation(a,a+9); } if(t<=0) puts("-1"); } return 0; }