A.2016
给出正整数 n 和 m，统计满足以下条件的正整数对 (a,b) 的数量：
1. 1≤a≤n,1≤b≤m;
2. a×b 是 2016 的倍数。
Input
输入包含不超过 30 组数据。
每组数据包含两个整数 n,m (1≤n,m≤10 9).
Output对于每组数据，输出一个整数表示满足条件的数量。Sample Input
32 63 2016 2016 1000000000 1000000000
Sample Output
1 30576 7523146895502644
代码：
#include<bits/stdc++.h> using namespace std; #define ll long long int vis[2020]; int main() { int n,m; while(~scanf("%d%d",&n,&m)) { memset(vis,0,sizeof(vis)); ll ans=0; for(int i=2016;i>=1;i--) { vis[i]+=n/i; for(int j=i-1;j>=1;j--) { if(i%j==0) vis[j]-=vis[i]; } ans+=1LL*vis[i]*(m/(2016/__gcd(2016,i))); } printf("%lld ",ans); } }
B. 有向无环图
Bobo 有一个 n 个点，m 条边的有向无环图（即对于任意点 v，不存在从点 v 开始、点 v 结束的路径）。
为了方便，点用 1,2,…,n 编号。 设 count(x,y) 表示点 x 到点 y 不同的路径数量（规定 count(x,x)=0），Bobo 想知道
除以 (10 9+7) 的余数。
其中，a i,b j 是给定的数列。
Input
输入包含不超过 15 组数据。
每组数据的第一行包含两个整数 n,m (1≤n,m≤10 5).
接下来 n 行的第 i 行包含两个整数 a i,b i (0≤a i,b i≤10 9).
最后 m 行的第 i 行包含两个整数 u i,v i，代表一条从点 u i 到 v i 的边 (1≤u i,vi≤n)。
Output对于每组数据，输出一个整数表示要求的值。Sample Input
3 3 1 1 1 1 1 1 1 2 1 3 2 3 2 2 1 0 0 2 1 2 1 2 2 1 500000000 0 0 500000000 1 2
Sample Output
4 4 250000014
代码：
#include<bits/stdc++.h> using namespace std; #define ll long long const int mod=1e9+7; const int maxn=1e5+10; ll a[maxn],b[maxn],s[maxn]; vector<int>G[maxn]; int in[maxn],n,m; void work() { queue<int>P; for(int i=1;i<=n;i++) { if(in[i]==0) P.push(i); } while(!P.empty()) { int v=P.front();P.pop(); for(int i=G[v].size()-1;i>=0;i--) { int u=G[v][i]; in[u]--; if(in[u]==0) P.push(u); s[u]=(s[u]+s[v])%mod; } } ll ans=0; for(int i=1;i<=n;i++) ans=(ans+(s[i]-a[i]+mod)*b[i])%mod; printf("%lld ",ans); } int main() { while(~scanf("%d%d",&n,&m)) { for(int i=1;i<=n;i++) { scanf("%lld%lld",&a[i],&b[i]); s[i]=a[i]; } for(int i=1;i<=m;i++) { int x,y;scanf("%d%d",&x,&y); G[x].push_back(y); in[y]++; } work(); for(int i=1;i<=n;i++) { G[i].clear(); in[i]=0; } } return 0; }
G.Parentthesis
Bobo has a balanced parenthesis sequence P=p 1 p 2…p n of length n and q questions.
The i-th question is whether P remains balanced after p ai and p bi swapped. Note that questions are individual so that they have no affect on others.
Parenthesis sequence S is balanced if and only if:
1. S is empty;
2. or there exists balanced parenthesis sequence A,B such that S=AB;
3. or there exists balanced parenthesis sequence S' such that S=(S').
Input
The input contains at most 30 sets. For each set:
The first line contains two integers n,q (2≤n≤10 5,1≤q≤10 5).
The second line contains n characters p 1 p 2…p n.
The i-th of the last q lines contains 2 integers a i,b i (1≤a i,b i≤n,a i≠b i).
OutputFor each question, output " Yes" if P remains balanced, or " No" otherwise.Sample Input
4 2 (()) 1 3 2 3 2 1 () 1 2
Sample Output
No Yes No
代码（括号匹配终于会啦）：
#include <bits/stdc++.h> using namespace std; const int maxn = 1e5 + 10; int N, Q; string s; int sum[maxn], vis[maxn], a[maxn]; int main() { while(~scanf("%d%d", &N, &Q)) { memset(sum, 0, sizeof(sum)); cin >> s; for(int i = 0; s[i]; i ++) { if(s[i] == '(') a[i] = 1; else a[i] = -1; } memset(vis, 0, sizeof(vis)); for(int i = 0; s[i]; i ++) { if(i == 0) sum[i] = a[i]; else sum[i] = sum[i - 1] + a[i]; } for(int q = 0; q < Q; q ++) { int l, r; scanf("%d%d", &l, &r); l -= 1, r -= 1; if(s[l] == s[r]) printf("Yes "); else { if(s[r] == ')') if(sum[l] - 2 < 0 || sum[r] - 2 < 0) printf("No "); else printf("Yes "); } } } return 0; }
H. Reverse
Bobo has a n digits decimal number D=d 1 d 2…d n (It may have leading zeros).
Let R(i,j) denotes number D with digits between the i-th position and j-th position reversed. That is, R(i,j)=d 1…d i-1 d j d j-1…d i d j+1 d j+2…d n.
Bobo would like to find
modulo (10 9+7).
Input
The input contains at most 30 sets. For each set:
The first line contains an integer n (1≤n≤10 5).
The second line contains n digits d 1 d 2…d n (0≤d i≤9).
OutputFor each set, an integer denotes the result.Sample Input
2 12 3 012 10 0123456789
Sample Output
45 369 733424314
代码：
#include<bits/stdc++.h> using namespace std; #define ll long long const int mod=1e9+7; ll pow1(ll a,ll b) { ll r=1; while(b) { if(b&1) r=r*a%mod; a=a*a%mod; b/=2; } return r; } ll inv_9=pow1(9,mod-2); ll p[100005]; char t[100005]; int main() { p[0]=1; for(int i=1;i<=100000;i++) p[i]=p[i-1]*10%mod; int n; while(~scanf("%d",&n)) { scanf("%s",t+1); reverse(t+1,t+n+1); ll ans=0; for(int i=1;i<=n;i++) { ll v=t[i]-'0'; ans+=(1LL*(i-1)+1LL*(i-1)*(i-2)/2)%mod*v%mod*p[i-1]%mod; ans+=(1LL*(n-i)+1LL*(n-i)*(n-i-1)/2)%mod*v%mod*p[i-1]%mod; ans%=mod; ll tmp=v*(p[i]-1+mod)%mod*inv_9%mod; tmp=tmp*(p[n-i+1]-1+mod)%mod*inv_9%mod; ans+=tmp; ans%=mod; } printf("%lld ",ans); } return 0; }
I.Tree Intersection
Bobo has a tree with n vertices numbered by 1,2,…,n and (n-1) edges. The i-th vertex has color c i, and the i-th edge connects vertices a i and b i.
Let C(x,y) denotes the set of colors in subtree rooted at vertex x deleting edge (x,y).
Bobo would like to know R_i which is the size of intersection of C(a i,b i) and C(bi,a i) for all 1≤i≤(n-1). (i.e. |C(a i,b i)∩C(b i,a i)|)
Input
The input contains at most 15 sets. For each set:
The first line contains an integer n (2≤n≤10 5).
The second line contains n integers c 1,c 2,…,c n (1≤c_i≤n).
The i-th of the last (n-1) lines contains 2 integers a i,b i (1≤a i,b i≤n).
OutputFor each set, (n-1) integers R 1,R 2,…,R n-1.Sample Input
4 1 2 2 1 1 2 2 3 3 4 5 1 1 2 1 2 1 3 2 3 3 5 4 5
Sample Output
1 2 1 1 1 2 1
代码：
#include<bits/stdc++.h> using namespace std; const int maxn=1e5+10; vector<int>G[maxn],id[maxn]; int a[maxn],ans[maxn],s[maxn],now[maxn],son[maxn],sz[maxn],cnt; void get_son(int v,int fa) { sz[v]=1; son[v]=0; for(int i=G[v].size()-1;i>=0;i--) { int u=G[v][i]; if(u==fa) continue; get_son(u,v); sz[v]+=sz[u]; if(sz[u]>sz[son[v]]) son[v]=u; } } void work1(int v,int fa) { now[a[v]]++; if(now[a[v]]==1&&s[a[v]]>1) cnt++; else if(now[a[v]]==s[a[v]]&&now[a[v]]>1) cnt--; for(int i=G[v].size()-1;i>=0;i--) { int u=G[v][i]; if(u==fa) continue; work1(u,v); } } void work2(int v,int fa) { now[a[v]]--; if(now[a[v]]==0&&s[a[v]]>1) cnt--; else if(now[a[v]]==s[a[v]]-1&&now[a[v]]>0) cnt++; for(int i=G[v].size()-1;i>=0;i--) { int u=G[v][i]; if(u==fa) continue; work2(u,v); } } void dfs(int v,int fa,int flag,int q) { int I; for(int i=G[v].size()-1;i>=0;i--) { int u=G[v][i]; if(u==son[v]) I=id[v][i]; if(u==fa||u==son[v]) continue; dfs(u,v,0,id[v][i]); } if(son[v]) dfs(son[v],v,1,I); for(int i=G[v].size()-1;i>=0;i--) { int u=G[v][i]; if(u==fa||u==son[v]) continue; work1(u,v); } now[a[v]]++; if(now[a[v]]==1&&s[a[v]]>1) cnt++; else if(now[a[v]]==s[a[v]]&&now[a[v]]>1) cnt--; ans[q]=cnt; if(flag==0) work2(v,fa); } int main() { int n; while(~scanf("%d",&n)) { memset(s,0,sizeof(s)); memset(now,0,sizeof(now)); for(int i=1;i<=n;i++) scanf("%d",&a[i]),s[a[i]]++; for(int i=1,x,y;i<n;i++) { scanf("%d%d",&x,&y); G[x].push_back(y); id[x].push_back(i); G[y].push_back(x); id[y].push_back(i); } get_son(1,0); cnt=0; dfs(1,0,0,0); for(int i=1;i<=n;i++) { G[i].clear(); id[i].clear(); if(i<n) printf("%d ",ans[i]); ans[i]=0; } } }
J.三角形和矩形
Bobo 有一个三角形和一个矩形，他想求他们交的面积。
具体地，三角形和矩形由 8 个整数 x 1,y 1,x 2,y 2,x 3,y 3,x 4,y 4 描述。 表示三角形的顶点坐标是 (x 1,y 1),(x 1,y 2),(x 2,y 1)， 矩形的顶点坐标是 (x 3,y 3),(x 3,y 4),(x 4,y4),(x 4,y 3).
Input
输入包含不超过 30000 组数据。
每组数据的第一行包含 4 个整数 x 1,y 1,x 2,y 2 (x 1≠x 2,y 1≠y 2).
第二行包含 4 个整数 x 3,y 3,x 4,y 4 (x 3<x 4,y 3<y 4).
(0≤x i,y i≤10 4)
Output对于每组数据，输出一个实数表示交的面积。绝对误差或相对误差小于 10 -6 即认为正确。Sample Input
1 1 3 3 0 0 2 2 0 3 3 1 0 0 2 2 4462 1420 2060 2969 4159 257 8787 2970
Sample Output
1.00000000 0.75000000 439744.13967527
代码：
#include<bits/stdc++.h> using namespace std; double k,b; double cal1(double Y1,double Y2,double y,double xx1,double xx2) { if(y>=Y2) return 0; double h=(y-b)/k; if(y>=Y1&&y<Y2) return (xx2-h)*(Y2-y)*0.5; return ((Y1-y)+(Y2-y))*(xx2-xx1)*0.5; } double cal2(double Y1,double Y2,double y,double xx1,double xx2) { if(y<=Y2) return 0; double h=(y-b)/k; if(y>Y2&&y<=Y1) return (y-Y2)*(xx2-h)*0.5; return ((y-Y1)+(y-Y2))*(xx2-xx1)*0.5; } double cal3(double Y1,double Y2,double y,double xx1,double xx2) { if(y>=Y1) return 0; double h=(y-b)/k; if(y>=Y2&&y<Y1) return (Y1-y)*(h-xx1)*0.5; return ((Y1-y)+(Y2-y))*(xx2-xx1)*0.5; } double cal4(double Y1,double Y2,double y,double xx1,double xx2) { if(y<=Y1) return 0; double h=(y-b)/k; if(y>Y1&&y<=Y2) return (y-Y1)*(h-xx1)*0.5; return ((y-Y1)+(y-Y2))*(xx2-xx1)*0.5; } int main() { int x1,y1,x2,y2,x3,y3,x4,y4; while(~scanf("%d%d%d%d%d%d%d%d",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4)) { int xx1=max(min(x1,x2),x3),xx2=min(max(x1,x2),x4); int yy1=max(min(y1,y2),y3),yy2=min(max(y1,y2),y4); k=1.0*(y2-y1)/(x1-x2),b=1.0*y2-k*x1; if(xx1>=xx2||yy1>=yy2) printf("0.0000000000000 "); else if(y2>y1&&x1>x2) { double Y1=k*xx1+b,Y2=k*xx2+b; double ans=cal1(Y1,Y2,1.0*yy1,1.0*xx1,1.0*xx2)-cal1(Y1,Y2,1.0*yy2,1.0*xx1,1.0*xx2); printf("%.10lf ",ans); } else if(y2<y1&&x2<x1) { double Y1=k*xx1+b,Y2=k*xx2+b; double ans=cal2(Y1,Y2,1.0*yy2,1.0*xx1,1.0*xx2)-cal2(Y1,Y2,1.0*yy1,1.0*xx1,1.0*xx2); printf("%.10lf ",ans); } else if(x2>x1&&y2>y1) { double Y1=k*xx1+b,Y2=k*xx2+b; double ans=cal3(Y1,Y2,1.0*yy1,1.0*xx1,1.0*xx2)-cal3(Y1,Y2,1.0*yy2,1.0*xx1,1.0*xx2); printf("%.10lf ",ans); } else { double Y1=k*xx1+b,Y2=k*xx2+b; double ans=cal4(Y1,Y2,1.0*yy2,1.0*xx1,1.0*xx2)-cal4(Y1,Y2,1.0*yy1,1.0*xx1,1.0*xx2); printf("%.10lf ",ans); } } }
明天省赛 加油啦 希望有好结果
今日份的瘦宅茶
喜茶最近出的芝芝桃桃和多肉粉荔都没时间去喝！！！不是合格的 HEYTEA Girl 了！！！快放假吧