打表。
#include<bits/stdc++.h>
#define mem(a,b) memset((a),b,sizeof(a))
#define de cout<<endl<<endl<<endl
typedef long long ll;
const int inf=0x3f3f3f3f;
const int N=100010;
using namespace std;
ll a[1000010][12];
ll solve(int n,int b)
{
ll ret=0;
while(n>0)
{
ret+=(n%b);
n/=b;
}
return ret;
}
int main()
{
for(int i=1;i<=1000000;i++)
for(int j=2;j<=10;j++)
a[i][j]=0;
for(int i=1;i<=1000000;i++)
for(int j=2;j<=10;j++)
a[i][j]=a[i-1][j]+solve(i,j);
int T;scanf("%d",&T);
int TT=0;
while(T--)
{
int n,b;
scanf("%d%d",&n,&b);
printf("Case #%d: %lld\n",++TT,a[n][b]);
}
return 0;
}思维题，不过不太好想。
#include<bits/stdc++.h>
#define mem(a,b) memset((a),b,sizeof(a))
#define de cout<<endl<<endl<<endl
typedef long long ll;
const int inf=0x3f3f3f3f;
const int N=100010;
using namespace std;
int a[2010];
int n,m;
int main()
{
int T;scanf("%d",&T);
int TT=0;
while(T--)
{
scanf("%d%d",&n,&m);
m*=2;
for(int i=1;i<=m;i++)
{
scanf("%d",&a[i]);
if(i%2==0)
a[i]++;
}
sort(a+1,a+m+1);
int ans=0;
for(int i=2;i<=m;i+=2)
ans+=(a[i]-a[i-1]);
printf("Case #%d: %d\n",++TT,ans);
}
return 0;
}倒着完全背包。
#include<bits/stdc++.h>
#define mem(a,b) memset((a),b,sizeof(a))
#define de cout<<endl<<endl<<endl
typedef long long ll;
const int inf=0x3f3f3f3f;
const int N=100010;
const ll mod=1e9+7;
using namespace std;
int n;
ll a[310],ans,sum;
ll dp[150010],p[310];
int main()
{
int T;scanf("%d",&T);
while(T--)
{
mem(dp,0);
mem(p,0);
dp[0]=1;
ans=0;
sum=0;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]),sum+=a[i];
sort(a+1,a+n+1);
p[n+1]=0;
for(int i=n;i>=1;i--)
p[i]=p[i+1]+a[i];
ll p1=(sum+1)/2,p2;
for(int i=n;i>=1;i--)
{
p2=(sum+a[i])/2;
for(int j=p1;j<=p2;j++)
{
if(j<a[i])
continue;
ans=(ans+dp[j-a[i]])%mod;
}
for(int j=p[i];j>=1;j--)
{
if(j-a[i]>=0)
dp[j]=(dp[j]+dp[j-a[i]])%mod;
else
break;
}
}
printf("%lld\n",ans);
}
return 0;
}参考博客：https://blog.youkuaiyun.com/yzsjwd/article/details/100867148
暴力打表。
打表代码：
#include<bits/stdc++.h>
#define mem(a,b) memset((a),b,sizeof(a))
#define de cout<<endl<<endl<<endl
typedef long long ll;
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
using namespace std;
ll f[3010],N,ans;
ll invf[3010],cnt[6010];
ll pow_mod(ll a,ll b)
{
ll ret=1;
while(b>0)
{
if(b&1)
ret=(ret*a)%mod;
a=(a*a)%mod;
b>>=1;
}
return ret;
}
void init()
{
f[0]=1;
f[1]=1;
for(int i=2; i<=3000; i++)
f[i]=f[i-1]*i%mod;
for(int i=1; i<=3000; i++)
invf[i]=pow_mod(f[i],mod-2);
}
/*
d：当前序列中非1的数的个数
p：序列中元素的最大值
n：非1元素最多有多少个
sum：当前和
prod：当前积
*/
void dfs(ll d,ll p,ll n,ll sum,ll prod)
{
ll t;
if(d==n) //非1的数的个数已经达到了
{
if(N>13)
t=N-13;
else
t=0;
if(sum+t==prod) //如果和等于积了
{
ll x=f[N];
x=(x*invf[cnt[1]+t])%mod;
for(int i=2;i<=6000;i++)
x=(x*invf[cnt[i]])%mod;
ans=(ans+x)%mod;
}
return ;
}
if(prod>6000)
return ;
for(int i=p;i<=6000/prod;i++)
{
cnt[i]++;
dfs(d+1,i,n,sum+i,prod*i);
cnt[i]--;
}
}
int main()
{
freopen("output.txt","w",stdout);
init();
f[0]=1;
invf[0]=1;
mem(cnt,0);
cout<<1;
for(N=2; N<=3000; N++)
{
ans=0;
ll t=min(N,1ll*13);
dfs(0,1,t,0,1);
cout<<","<<ans;
}
return 0;
}AC代码：
#include<bits/stdc++.h>
#define mem(a,b) memset((a),b,sizeof(a))
#define de cout<<endl<<endl<<endl
typedef long long ll;
const int inf=0x3f3f3f3f;
const int N=100010;
using namespace std;
ll 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3,644072860,608828254,679830843,408364825,819994273,708409577,33289493,729955094,596908384,552676290,804063672,642210283,87969249,139856655,573837082,585204162,709919080,62019450,690340904,134375538,239873992,248897451,344866625,929545102,553885732,378658393,844885049,839814934,567317169,38067232,429000062,781481111,237895107,782073191,337473894,141506057,492977927,460288582,745195402,482110822,102004165,152293447,685882358,485879386,848080146,599355957,488608364,19053402,423318689,471333487,188166164,302107539,372743633,177700432,978971079,366701808,401596902,927509072,171361096,95127688,424301456,399542853,768237456,601860578,499990629,187805495,191279700,481600043,305424704,222429969,602590712,904063876,842352983,944208736,819657360,49463685,538026209,906352779,419674963,930518965,778141124,255585286,33994747,519787112,79986460,226108045,524476702,27656221,255579862,646238174,847099521,272965646,865493485,385727148,126314306,37843618,788938122,618363112,928198145,400264960,871653169,62240715,364773082,617467205,972975522,710291243,636653364,440799981,141210740,57867049,423646746,714493511,177625722,617783970,274650380,687499716,611743818,664745550,378779532,204366492,162895731,549949937,442481335,525455322,740477352,481424096,305714815,51047601,438699099,262169947,133720731,478955102,436063979,595492260,936083246,863122159,945175103,263539607,510732407,282342468,686191734,492765857,436266853,1776510,904247837,612920718,48007182,239884386,11204682,40517916,620896715,428051563,228035641,81480630,987941217,506734291,827420210,315533927,873321075,194012513,701591193,469021066,977224823,287810653,418116475,229914328,680362293,685870845,221377050,217438428,436976820,956610315,609746286,995059943,834130154,620333413,129424212,85123869,684341657,372232483,849850868,39008962,334692007,441172896,441332548,832399170,559399752,817223339,368656463,508150161,982500386,805756404,361093783,782516904,227697402,552864398,376648315,131219359,824310464,88378402,105093721,931111394,978424570,462045876,775812495,375392919,974899300,472606090,544185580,50284084,401130151,414301598,593630705,917370510,658630453,411490799,178655844,380745385,325174122,3082222,682403282,550267455,841077458,610783619,361233023,793693746,460107087,799438353,760107599,52044074,119468097,44303927,132339297,573405675,272388138,43419227,411058183,220220432,611745919,805350568,970746014,764062232,393882739,983284303,272507287,577270472,846617654,491819287,206119318,521558324,910021282,639310249,620356611,64059745,687961292,625991240,25907400,59892214,88040640,776895239,818968695,160783525,954869542,293693723,468987937,446082996,916942884,311292313,658592565,634756596,70358162,210975824,253238476,832920931,963525669,882776534,763829888,717755102,134562859,992220619,532109146,425961811,493677646,107786653,207818522,196066339,111474900,257644627,274562332,307727812,717836315,326865766,873090201,124492385,846032717,722268565,757127484,332116994,432329903,443274582,610441402,698528081,310044953,260046011,554112619,703742007,402774419,292306979,760774539,402066147,752998364,271696960,177313912,309878494,540104932,779043454,569185053,160233288,325474984,756301940,892820990,500053160,231334896,794606267,404678875,980732369,283755839,368386613,86296858,67762572,75074675,991932675,688983358,913104550,786957021,879078663,146575829,191767008,638549864,146213572,79225954,65949125,267399617,616601372,31296380,534786460,871570500,417735380,771785110,376385944,154401309,234734748,631694290,542886380,102226975,785537721,242796494,777292172,650777687,121632024,490436396,331066103,939668693,433716263,980024850,797678562,305812176,982021027,688100320,829876446,125002285,364614521,219921143,245211624,309829642,43455167,364536486,85737790,425628595,859372492,471784346,403385340,96571820,837277437,386811318,745690100,28585286,438708382,6635488,915376438,487080701,247830110,935059609,76336473,816549921,272103268,768416972,859456661,463819251,32608561,698776266,688501880,939605786,5192087,844383395,550698387,396220525,25232358,122324978,874042807,493690026,447201459,855822480,585455069,894491961,989605241,539114271,273794598,517458429,753561258,41854924,936179443,534463507,784177424,645763670,839030396,385673367,619137125,536106422};
int main()
{
int T;scanf("%d",&T);
int n;
while(T--)
{
scanf("%d",&n);
printf("%d\n",ans[n]);
}
return 0;
}题意：求出合法的长度为n的字典序第k小字符串，合法的定义为除了最后一位，每一位的取值范围为'A'到'A'+pos-1，而最后一位的取值范围'A'到当前字符串最大值+1。
思路：dp求每个节点下面有多少种情况，如果此节点下的情况数大于等于k，那就输出这个节点，并且k-=这个节点下的情况数；否则，继续判断下一个节点。dp[n][i][j]表示输入的数为n时，第 i 层中某个节点下有多少种情况，并且这个节点需要满足的要求为这条线路上最大的字符为j（如果最大字符为‘A'，那么j等1，如果最大字符为’Z'，那么j等26）。
dp[i][j]=dp[i+1][j]*j+dp[i+1][j+1]。初始化dp[i][i][j]=1。i为1到26，j为1到i。详见代码。
因为当n等26的时候，dp会很大，long long存不下，所以要用到__int128，输出的话，必须用快读。听说用java会超时。
ACcode:20ms
#include<bits/stdc++.h>
#define mem(a,b) memset((a),b,sizeof(a))
#define de cout<<endl<<endl<<endl
#define ll __int128
//#define ll long long
const int inf=0x3f3f3f3f;
const int N=100010;
using namespace std;
ll n,k;
ll dp[30][30][30];
void init()
{
mem(dp,0);
for(int i=1;i<=26;i++)
{
for(int j=1;j<=i;j++)
dp[i][i][j]=1;
for(int j=i-1;j>=1;j--)
{
for(int k=1;k<=j;k++)
dp[i][j][k]=dp[i][j+1][k]*k+dp[i][j+1][k+1];
}
}
}
inline __int128 read()
{
__int128 x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9')
{
if(ch=='-')
f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
x=x*10+ch-'0';
ch=getchar();
}
return x*f;
}
//inline void print(__int128 x)
//{
// if(x<0)
// {
// putchar('-');
// x=-x;
// }
// if(x>9)
// print(x/10);
// putchar(x%10+'0');
//}
int main()
{
init();
int T;scanf("%d",&T);
int TT=0;
while(T--)
{
//scanf("%lld%lld",&n,&k);
//read(n);read(k);
n=read();k=read();
printf("Case #%d: ",++TT);
int p=1,mx;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=i;j++)
{
mx=max(p,j);
if(dp[n][i][mx]>=k)
{printf("%c",char(j+'A'-1));p=max(p,j);break;}
else
k-=dp[n][i][mx];
}
}
printf("\n");
}
return 0;
}题意：输入n，然后给你三个长度均为n的数组a、b、c，问你存在多少对i、j、k满足|Ai-Bj|<=C[k]，且|Bj-Ck|<=Ai，且|Ai-Ck|<=Bj。
思路：FFT+容斥。
实际上题意可以转化一下，问的就是存在多少对ijk满足两边之和大于等于第三边，也就是满足“能构成三角形”或者“两边之和等于第三边”。那我们可以容斥一下，答案就是n*n*n-“有多少种情况满足任意两边之和小于第三边”。至于“有多少种情况满足任意两边之和小于第三边”我们可以FFT求。
还有，FFT的时间复杂度是O(n*longn)。当n小于等于1000的时候，可以直接O(n*n)暴力跑一下；如果n大于1000，再用FFT。至于为什么不能直接用FFT，好像是因为FFT的复杂度有一个很大的常数，当n小于等于1000的时候，O(n*n)是小于O(n*logn)+常数的。这个题还比较坑，就是如果你写成n小于等于1000直接O(n*n)就能过，但是如果写成n小于1000直接O(n*n)就会T......应该是因为有组n等1000的数据等着卡你。
ACcode：1986ms。
#include<bits/stdc++.h>
#define mem(a,b) memset((a),b,sizeof(a))
#define de cout<<endl<<endl<<endl
typedef long long ll;
const int inf=0x3f3f3f3f;
using namespace std;
ll N;
ll A[100010],B[100010],C[100010];
ll suma[100010],sumb[100010],sumc[100010];
#define maxn 400005//开四倍所有数组。
const double pi=acos(-1.0);
struct Complex
{
double x,y;
Complex(double X=0,double Y=0)
{
x=X,y=Y;
}
} a[maxn],b[maxn];
Complex operator + (Complex a,Complex b)
{
return Complex(a.x+b.x,a.y+b.y);
}
Complex operator - (Complex a,Complex b)
{
return Complex(a.x-b.x,a.y-b.y);
}
Complex operator * (Complex a,Complex b)
{
return Complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
}
int S,T,n,m,L,R[maxn];
long long F[maxn];//存储多项式相乘结果的系数从0-n
double f[maxn],g[maxn];
void FFT(Complex a[maxn],int opt)
{
for (int i=0; i<n; ++i)
if (i<R[i]) swap(a[i],a[R[i]]);
for (int k=1; k<n; k<<=1)
{
Complex wn=Complex(cos(pi/k),opt*sin(pi/k));
for (int i=0; i<n; i+=(k<<1))
{
Complex w=Complex(1,0);
for (int j=0; j<k; ++j,w=w*wn)
{
Complex x=a[i+j],y=w*a[i+j+k];
a[i+j]=x+y,a[i+j+k]=x-y;
}
}
}
}
void calc(int opt)
{
FFT(a,1);
FFT(b,1);
for (int i=0; i<=n; ++i)
a[i]=a[i]*b[i];
FFT(a,-1);
for (int i=0; i<=n; ++i)//上限为n，在运用时灵活修改。
F[i]=(long long)(a[i].x/n+0.5)*opt;
}
void solve()
{
/*
多项式f和多项式g的乘积
其中，多项式f的次数为0到S，多项式g的次数为0到T
结果存在数组F中，F的系数为0到S+T
时间：O(n logn)
*/
m=S+T;
L=0;//多组输入时，L要初始化为0
for (n=1; n<=m; n<<=1) ++L;
for (int i=0; i<n; ++i)
R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
for (int i=0; i<=n; ++i)
a[i]=Complex(1.0*f[i],0.0),b[i]=Complex(1.0*g[i],0.0);
calc(1);
}
void init()
{
mem(f,0);
mem(g,0);
mem(F,0);
}
int main()
{
int T1;scanf("%d",&T1);
int TT=0;
while(T1--)
{
init();
scanf("%lld",&N);
for(int i=1;i<=N;i++)
scanf("%lld",&A[i]);
for(int i=1;i<=N;i++)
scanf("%lld",&B[i]);
for(int i=1;i<=N;i++)
//C[i]=read();
scanf("%lld",&C[i]);
sort(A+1,A+N+1);
sort(B+1,B+N+1);
sort(C+1,C+N+1);
mem(suma,0);
mem(sumb,0);
mem(sumc,0);
for(int i=1;i<=N;i++)
suma[A[i]]++;
for(int i=2;i<=A[N];i++)
suma[i]=suma[i-1]+suma[i];
for(int i=1;i<=N;i++)
sumb[B[i]]++;
for(int i=2;i<=B[N];i++)
sumb[i]=sumb[i-1]+sumb[i];
for(int i=1;i<=N;i++)
sumc[C[i]]++;
for(int i=2;i<=C[N];i++)
sumc[i]=sumc[i-1]+sumc[i];
ll ans=0;
if(N<=2000)
{
for(int i=1;i<=N;i++)
for(int j=1;j<=N;j++)
{
if(A[i]+B[j]>C[N])
break;
ans+=(N-sumc[A[i]+B[j]]);
}
for(int i=1;i<=N;i++)
for(int j=1;j<=N;j++)
{
if(A[i]+C[j]>B[N])
break;
ans+=(N-sumb[A[i]+C[j]]);
}
for(int i=1;i<=N;i++)
for(int j=1;j<=N;j++)
{
if(B[i]+C[j]>A[N])
break;
ans+=(N-suma[B[i]+C[j]]);
}
printf("Case #%d: ",++TT);
ans=N*N*N-ans;
printf("%lld\n",ans);
continue;
}
S=A[N] + 10;
T=B[N] + 10;
n=S+T+10;
for(int i=1;i<=N;i++)
f[A[i]]++;
for(int i=1;i<=N;i++)
g[B[i]]++;
solve();
for(int i=2;i<=n;i++)
{
if(i>C[N])
break;
ans+=(F[i]*(N-sumc[i]));
}
init();
S=A[N];
T=C[N];
n=S+T;
for(int i=1;i<=N;i++)
f[A[i]]++;
for(int i=1;i<=N;i++)
g[C[i]]++;
solve();
for(int i=2;i<=n;i++)
{
if(i>B[N])
break;
ans+=(F[i]*(N-sumb[i]));
}
init();
S=C[N];
T=B[N];
n=S+T;
for(int i=1;i<=N;i++)
f[C[i]]++;
for(int i=1;i<=N;i++)
g[B[i]]++;
solve();
for(int i=2;i<=n;i++)
{
if(i>A[N])
break;
ans+=(F[i]*(N-suma[i]));
}
printf("Case #%d: ",++TT);
ans=N*N*N-ans;
printf("%lld\n",ans);
}
return 0;
}题意：让你构造一个长度为n的序列满足：1、序列中每个元素的值都在1到m之间；2、对于序列中任意一个偶数，需要满足它出现的次数是偶数。
思路：推公式，然后预处理，然后O(1)求组合数。
至于公式怎么退，我不会......
有篇感觉讲的很不错的博客，可我还是不会。https://www.cnblogs.com/heyuhhh/p/11545443.html
还有，这个题有个我不知道的地方就是，对于除法，除完后要取模，比如ans=a/b%mod，不能直接写成ans=a/b%mod，应该写成ans=a*pow_mod(b,mod-2,mod)%mod，也就是说要写成逆元的形式。
推公式的过程：
ACcode：148ms
#include<bits/stdc++.h>
#define mem(a,b) memset((a),b,sizeof(a))
#define de cout<<endl<<endl<<endl
typedef long long ll;
typedef long long LL;
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
using namespace std;
ll n,m;
ll pow_mod(ll a,ll b)
{
ll ret=1;
while(b)
{
if(b&1)
ret=ret*a%mod;
b>>=1;
a=a*a%mod;
}
return ret;
}
const int maxn = 2e5;
LL A[maxn];
LL B[maxn];
void Init()
{
A[0] = 1;
for(int i=1; i<=maxn; i++)
A[i] = (A[i-1] * i ) % mod;
}
LL Ext_Gcd(LL a, LL b, LL &x, LL &y)
{
if(b==0){
x=1;
y=0;
return a;
}
LL d = Ext_Gcd(b, a%b, y, x);
y-=a/b*x;
return d;
}
LL Inv(LL a, LL n)
{
LL x,y;
LL d = Ext_Gcd(a,n,x,y);
if(d == 1)
return ((x%n)+n)%n;
return -1;
}
LL get()
{
for(int i=0;i<maxn;i++)
B[i] = Inv(A[i],mod);
}
LL C(LL a, LL b, LL mod)
{
if(a < b)
return 0;
return (A[a] * B[b] %mod)*B[a-b]%mod;
}
inline LL mmp(int x)
{
if(x & 1)
return -1;
else
return 1;
}
int main()
{
Init();
get();
int T;scanf("%d",&T);
while(T--)
{
scanf("%lld%lld",&n,&m);
ll cnt0=m/2,cnt1=(m+1)/2;
ll ans=0;
for(int i=0;i<=cnt0;i++)
ans=(ans+C(cnt0,1ll*i,mod)*pow_mod(1ll*(cnt1-cnt0+2*i),n)%mod)%mod;
//ans=ans/pow_mod(1ll*2,cnt0)%mod;
//不能这么写，要用逆元写
ll anss=pow_mod(1ll*2,cnt0);
ans=(ans*pow_mod(anss,mod-2))%mod;
printf("%lld\n",ans);
}
return 0;
}其他的真不会了，以后再说吧......
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