题面：https://www.cnblogs.com/Juve/articles/11767239.html
94，95的T3都没改出来，是我太菜了。。。
凉宫春日的忧郁：
比较$x^y$和$y!$的大小，如果打高精会T掉
正解：把两个数取log，则$log_2x^y=ylog_2x$，$log_2y!=sumlimits_{i=1}^{y}log_2i$
然后就A了
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 #define int long long 7 using namespace std; 8 int t,x,y; 9 signed main(){ 10 freopen("yuuutsu.in","r",stdin); 11 freopen("yuuutsu.out","w",stdout); 12 scanf("%lld",&t); 13 while(t--){ 14 scanf("%lld%lld",&x,&y); 15 long double xx=y*log2(x); 16 long double yy=0.0; 17 for(int i=1;i<=y;++i){ 18 yy+=log2(i); 19 } 20 if(xx<=yy) puts("Yes"); 21 else puts("No"); 22 } 23 return 0; 24 }
漫无止境的八月：
如果满足的话必须保证所有%k同余的位置上的和一样，所以打个map就好了。。。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<unordered_map> 6 using namespace std; 7 const int MAXN=2e6+5; 8 int read(){ 9 int x=0,f=1;char ch=getchar(); 10 while(ch<'0'||ch>'9'){ 11 if(ch=='-') f=-1; 12 ch=getchar(); 13 } 14 while(ch>='0'&&ch<='9'){ 15 x=(x<<3)+(x<<1)+ch-'0'; 16 ch=getchar(); 17 } 18 return x*f; 19 } 20 int n,k,q,a[MAXN],sum[MAXN]; 21 unordered_map<int,int>mp; 22 signed main(){ 23 freopen("august.in","r",stdin); 24 freopen("august.out","w",stdout); 25 n=read(),k=read(),q=read(); 26 for(int i=1;i<=n;++i){ 27 a[i]=read(); 28 sum[i%k]+=a[i]; 29 } 30 for(int i=0;i<k;++i) ++mp[sum[i%k]]; 31 if(mp[sum[0]]==k) puts("Yes"); 32 else puts("No"); 33 for(int i=1;i<=q;++i){ 34 int pos=read(),val=read(); 35 a[pos]+=val; 36 --mp[sum[pos%k]]; 37 sum[pos%k]+=val; 38 ++mp[sum[pos%k]]; 39 if(mp[sum[0]]==k) puts("Yes"); 40 else puts("No"); 41 } 42 return 0; 43 } 44 /* 45 5 2 5 46 1 1 1 2 1 47 3 −1 48 1 −1 49 3 1 50 3 1 51 1 −1 52 */
简单计算：
$2*sumlimits_{i=0}^{p}lfloorfrac{i*q}{p} floor=sumlimits_{i=0}^{p}lfloorfrac{i*q}{p} floor+lfloorfrac{(p-i)*q}{p} floor$
所以原式=$(p+1)*q-sumlimits_{i=0}^{p}[(p|i*q)?0:1]=(p+1)*q-p+gcd(p,q)$
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #define int long long 6 using namespace std; 7 int t,p,q,ans; 8 int gcd(int a,int b){ 9 return b==0?a:gcd(b,a%b); 10 } 11 signed main(){ 12 freopen("simplecalc.in","r",stdin); 13 freopen("simplecalc.out","w",stdout); 14 scanf("%lld",&t); 15 while(t--){ 16 scanf("%lld%lld",&p,&q); 17 ans=(p+1)*q-p+gcd(p,q); 18 printf("%lld ",ans>>1); 19 } 20 return 0; 21 }
格式化：
一个贪心，肯定是先选对容量有贡献的，即格式化后容量增加的，再选容量不增的，再选容量减少的，对于容量增加的，内部按格式化前从小到大排序，对于容量减小的，内部按格式化后的从大到小排序，然后check即可
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #define int long long 6 using namespace std; 7 const int MAXN=1e6+5; 8 int n,ans=0x3f3f3f3f3f3f3f3f,l,r; 9 struct node{ 10 int pre,now,w; 11 friend bool operator < (node p,node q){ 12 if(p.w>0&&q.w>0){ 13 return p.pre<q.pre; 14 } 15 if(p.w<0&&q.w<0){ 16 return p.now>q.now; 17 } 18 if(p.w==0&&q.w==0){ 19 return p.pre>q.pre; 20 } 21 if(p.w==0){ 22 return q.w<0; 23 } 24 if(q.w==0){ 25 return p.w>0; 26 } 27 if(p.w>0&&q.w<0) return 1; 28 if(p.w<0&&q.w>0) return 0; 29 return 1; 30 } 31 }a[MAXN]; 32 bool check(int val){ 33 for(int i=1;i<=n;++i){ 34 if(a[i].pre>val) return 0; 35 val-=a[i].pre,val+=a[i].now; 36 } 37 return 1; 38 } 39 signed main(){ 40 freopen("reformat.in","r",stdin); 41 freopen("reformat.out","w",stdout); 42 scanf("%lld",&n); 43 for(int i=1;i<=n;++i){ 44 scanf("%lld%lld",&a[i].pre,&a[i].now); 45 r+=a[i].pre; 46 a[i].w=a[i].now-a[i].pre; 47 } 48 sort(a+1,a+n+1); 49 while(l<r){ 50 int mid=(l+r)>>1; 51 if(check(mid)) ans=min(ans,mid),r=mid; 52 else l=mid+1; 53 } 54 printf("%lld ",ans); 55 return 0; 56 }
真相：
我好弱啊，我太菜了
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 const int MAXN=1e6+5; 8 int t,n,sta[MAXN],top=0,tru[MAXN],sum0[MAXN],sum1[MAXN],f[MAXN],g[MAXN]; 9 struct node{ 10 int opt,val; 11 }a[MAXN]; 12 bool flagg=0; 13 int calc(int p){ 14 return (p-1!=0)?(p-1):n; 15 } 16 vector<int>v[MAXN]; 17 signed main(){ 18 freopen("truth.in","r",stdin); 19 freopen("truth.out","w",stdout); 20 scanf("%d",&t); 21 while(t--){ 22 top=flagg=0; 23 scanf("%d",&n); 24 for(int i=1;i<=n;++i){ 25 char op[2]; 26 scanf("%s",op); 27 if(op[0]=='+'){ 28 a[i].opt=1; 29 }else if(op[0]=='-'){ 30 a[i].opt=2; 31 }else{ 32 a[i].opt=3; 33 scanf("%d",&a[i].val); 34 flagg=1; 35 sta[++top]=i; 36 } 37 } 38 if(!flagg){ 39 bool now=0; 40 for(int i=1;i<=n;++i){ 41 if(now==0){ 42 if(a[i].opt==1){ 43 now=0; 44 }else now=1; 45 }else{ 46 if(a[i].opt==1){ 47 now=1; 48 }else now=0; 49 } 50 } 51 if(now==0) puts("consistent"); 52 else puts("inconsistent"); 53 continue; 54 }else{ 55 bool flag=0; 56 for(int i=1;i<=top;++i){ 57 int yy=sta[i]; 58 ++sum1[yy]; 59 int p=yy; 60 bool now=1; 61 while(a[calc(p)].opt!=3){ 62 p=calc(p); 63 if(now==0){ 64 if(a[p].opt==2) now=1; 65 else now=0; 66 }else{ 67 if(a[p].opt==1) now=1; 68 else now=0; 69 } 70 sum1[yy]+=now; 71 } 72 p=yy; 73 now=0; 74 while(a[calc(p)].opt!=3){ 75 p=calc(p); 76 if(now==0){ 77 if(a[p].opt==2) now=1; 78 else now=0; 79 }else{ 80 if(a[p].opt==1) now=1; 81 else now=0; 82 } 83 sum0[yy]+=now; 84 } 85 } 86 int num=0; 87 for(int i=1;i<=top;++i){ 88 f[a[sta[i]].val]+=sum0[sta[i]]; 89 g[a[sta[i]].val]+=sum1[sta[i]]; 90 num+=sum0[sta[i]]; 91 } 92 for(int i=0;i<=n;++i){ 93 num-=f[i]; 94 num+=g[i]; 95 if(num==i){ 96 flag=1; 97 break; 98 } 99 num-=g[i]; 100 num+=f[i]; 101 } 102 if(flag) puts("consistent"); 103 else puts("inconsistent"); 104 for(int i=1;i<=top;++i){ 105 f[a[sta[i]].val]=g[a[sta[i]].val]=0; 106 sum0[sta[i]]=sum1[sta[i]]=0; 107 } 108 } 109 } 110 return 0; 111 } 112 /* 113 1 114 3 115 $ 0 116 - 117 - 118 119 */