T1 任务安排
容易发现 ans一定在0到min(s[i],t[i]) 的范围内，二分这个最早时间，按着完成工作、
1 #include <algorithm> 2 #include <cstdio> 3 4 inline void read(int &x) 5 { 6 x=0; register char ch=getchar(); 7 for(; ch>'9'||ch<'0'; ) ch=getchar(); 8 for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0'; 9 } 10 const int N(100005); 11 struct Work { 12 int t,s; 13 bool operator < (const Work&x)const 14 { 15 if(s==x.s) return t<x.t; 16 return s<x.s; 17 } 18 }job[N]; 19 20 int ans=-1,L,R,Mid,n; 21 inline bool check(int x) 22 { 23 for(int i=1; i<=n; ++i) 24 { 25 if(x+job[i].t>job[i].s) return 0; 26 x+=job[i].t; 27 } 28 return 1; 29 } 30 31 int Presist() 32 { 33 freopen("manage.in","r",stdin); 34 freopen("manage.out","w",stdout); 35 36 read(n); 37 for(int t,i=1; i<=n; ++i) 38 read(job[i].t),read(job[i].s); 39 std::sort(job+1,job+n+1); 40 for(R=job[1].s-job[1].t+1; L<=R; ) 41 { 42 Mid=L+R>>1; 43 if(check(Mid)) 44 { 45 ans=Mid; 46 L=Mid+1; 47 } 48 else R=Mid-1; 49 } 50 printf("%d ",ans); 51 return 0; 52 } 53 54 int Aptal=Presist(); 55 int main(int argc,char**argv){;}
貌似可以直接贪心过。。
T2 小 a 的强迫症
处理个数前缀和sum[i],对于第i个珠子对答案的贡献为C(sum[i]-1,num[i]-1)，乘法原理统计ans
1 #include <cstdio> 2 3 #define LL long long 4 5 inline void read(int &x) 6 { 7 x=0; register char ch=getchar(); 8 for(; ch>'9'||ch<'0'; ) ch=getchar(); 9 for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0'; 10 } 11 const int mod(998244353); 12 const int MAX(500005); 13 const int N(100005); 14 int num[N],sum[N]; 15 LL jc[MAX]; 16 17 inline LL Pow(LL a,LL b) 18 { 19 LL ret=1; 20 for(; b; b>>=1,a=(a%mod)*(a%mod)%mod) 21 if(b&1) ret=(ret%mod)*(a%mod)%mod; 22 return ret%mod; 23 } 24 25 inline LL C(int a,int b) 26 { 27 return jc[a]%mod*Pow(jc[b],mod-2)%mod*Pow(jc[a-b],mod-2)%mod; 28 } 29 30 int Presist() 31 { 32 freopen("qiang.in","r",stdin); 33 freopen("qiang.out","w",stdout); 34 35 int n; read(n); jc[0]=1; 36 for(int i=1; i<=n; ++i) 37 read(num[i]),sum[i]=sum[i-1]+num[i]; 38 for(int i=1; i<=sum[n]; ++i) 39 jc[i]=jc[i-1]%mod*i%mod%mod; 40 LL ans=1; 41 for(int i=2; i<=n; ++i) 42 ans=ans%mod*C(sum[i]-1,num[i]-1)%mod; 43 printf("%I64d ",ans); 44 return 0; 45 } 46 47 int Aptal=Presist(); 48 int main(int argc,char**argv){;}
T3 函数求和
1 #include <cstdio> 2 3 #define LL long long 4 5 inline void read(int &x) 6 { 7 x=0; register char ch=getchar(); 8 for(; ch>'9'||ch<'0'; ) ch=getchar(); 9 for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0'; 10 } 11 const int N(1e5+5); 12 int a[N],L[N],R[N],sum[N]; 13 14 struct Tree { 15 int l,r; 16 LL sum; 17 bool flag; 18 }tr[2][N<<2]; 19 20 #define lc (now<<1) 21 #define rc (now<<1|1) 22 23 void Tree_push(int now) 24 { 25 tr[0][lc].flag+=tr[0][now].flag; 26 tr[0][lc].sum+=tr[0][now].flag; 27 tr[0][rc].flag+=tr[0][now].flag; 28 tr[0][rc].sum+=tr[0][now].flag; 29 tr[0][now].flag=0; 30 } 31 32 void Tree_change(int now,int to,int x) 33 { 34 if(tr[0][now].l==tr[0][now].r) 35 { 36 tr[0][now].sum=x; 37 tr[0][now].flag+=x; 38 return ; 39 } 40 if(tr[0][now].flag) Tree_push(now); 41 int mid=tr[0][now].l+tr[0][now].r>>1; 42 if(to<=mid) Tree_change(lc,to,x); 43 else if(to>mid) Tree_change(rc,to,x); 44 tr[0][now].sum=tr[0][lc].sum+tr[0][rc].sum; 45 } 46 47 LL Tree_query(int now,int l,int r,int op) 48 { 49 if(tr[op][now].l==l&&tr[op][now].r==r) return tr[op][now].sum; 50 int mid=tr[op][now].l+tr[op][now].r>>1; 51 if(r<=mid) return Tree_query(lc,l,r,op); 52 else if(l>mid) return Tree_query(rc,l,r,op); 53 else return Tree_query(lc,l,mid,op)+Tree_query(rc,mid+1,r,op); 54 } 55 56 void Tree_build(int now,int l,int r,int op) 57 { 58 tr[op][now].l=l,tr[op][now].r=r; 59 if(l==r) 60 { 61 if(!op) tr[op][now].sum=a[l]; 62 else tr[op][now].sum=Tree_query(1,L[l],R[l],0); 63 tr[op][now].flag=0; return ; 64 } 65 int mid=tr[op][now].l+tr[op][now].r>>1; 66 Tree_build(lc,l,mid,op); Tree_build(rc,mid+1,r,op); 67 tr[op][now].sum=tr[op][lc].sum+tr[op][rc].sum; 68 } 69 70 int Presist() 71 { 72 freopen("sum.in","r",stdin); 73 freopen("sum.out","w",stdout); 74 int n; read(n); 75 for(int i=1; i<=n; ++i) read(a[i]); 76 Tree_build(1,1,n,0); 77 for(int i=1; i<=n; ++i) 78 read(L[i]),read(R[i]); 79 Tree_build(1,1,n,1); 80 int q; read(q); 81 for(int op,l,r; q--; ) 82 { 83 read(op),read(l),read(r); 84 if(op==1) 85 { 86 Tree_change(1,l,r); 87 Tree_build(1,1,n,1); 88 } 89 else printf("%I64d ",Tree_query(1,l,r,1)); 90 } 91 return 0; 92 } 93 94 int Aptal=Presist(); 95 int main(int argc,char**argv){;}
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<cmath> 5 using namespace std; 6 typedef unsigned long long LL; 7 #define lowbit(x) ((x)&-(x)) 8 int n; 9 int a[100005]; 10 LL c[100005]; 11 int L[100005],R[100005]; 12 int Q; 13 int t[405][100005]; 14 int belong[100005];// (i-1)/B+1 15 LL sum[405]; 16 int B,NUM; 17 void add(int x,LL d){ 18 while(x<=n){ 19 c[x]+=d; 20 x+=lowbit(x); 21 } 22 } 23 LL ask(int x){ 24 LL ret=0; 25 while(x){ 26 ret+=c[x]; 27 x-=lowbit(x); 28 } 29 return ret; 30 } 31 int main(){ 32 freopen("sum.in","r",stdin); 33 freopen("sum.out","w",stdout); 34 scanf("%d",&n); 35 for(int i=1;i<=n;i++){ 36 scanf("%d",&a[i]); 37 } 38 for(int i=1;i<=n;i++){ 39 scanf("%d%d",&L[i],&R[i]); 40 } 41 B=sqrt(n)+1; 42 // printf("#%d ",B); 43 NUM=(n-1)/B+1; 44 for(int i=1;i<=n;i++) belong[i]=(i-1)/B+1; 45 for(int i=1;i<=n;i++) c[i]=a[i]; 46 for(int i=1;i<=n;i++){ 47 if(i+lowbit(i)<=n){ 48 c[i+lowbit(i)]+=c[i]; 49 } 50 } 51 for(int i=1;i<=n;i++){ 52 t[belong[i]][L[i]]++; 53 t[belong[i]][R[i]+1]--; 54 } 55 for(int i=1;i<=NUM;i++){ 56 for(int j=1;j<=n;j++){ 57 t[i][j]+=t[i][j-1]; 58 sum[i]+=t[i][j]*1ULL*a[j]; 59 } 60 } 61 scanf("%d",&Q); 62 while(Q--){ 63 int type,x,y; 64 scanf("%d%d%d",&type,&x,&y); 65 if(type==1){ 66 for(int i=1;i<=NUM;i++){ 67 sum[i]-=t[i][x]*1ULL*a[x]; 68 sum[i]+=t[i][x]*1ULL*y; 69 } 70 add(x,-a[x]); 71 a[x]=y; 72 add(x,a[x]); 73 }else{ 74 int Ln,Rn; 75 Ln=belong[x],Rn=belong[y]; 76 LL ans=0; 77 if(Ln==Rn){ 78 for(int i=x;i<=y;i++){ 79 ans+=ask(R[i])-ask(L[i]-1); 80 } 81 }else{ 82 for(int i=Ln+1;i<Rn;i++) ans+=sum[i]; 83 int lim; 84 lim=Ln*B; 85 lim=min(lim,y); 86 for(int i=x;i<=lim;i++){ 87 ans+=ask(R[i])-ask(L[i]-1); 88 } 89 lim=(Rn-1)*B+1; 90 lim=max(lim,x); 91 for(int i=lim;i<=y;i++){ 92 ans+=ask(R[i])-ask(L[i]-1); 93 } 94 } 95 printf("%I64d ",ans); 96 } 97 } 98 fclose(stdout); 99 return 0; 100 }