题解

36 条题解

  • 2
    @ 2017-11-06 07:46:30

    杨辉三角加快速幂秒过 (o ° ω ° O )
    记得用long long!!!被这个卡了五次

    #include<iostream>
    using namespace std;
    long long a,b,k,n,m,ans;
    long long yh[1005][1005];
    int quickpow(int a,int b,int c)
    {
        int ans=1;
        a=a%c;
        while(b!=0)
        {
            if(b&1)
            ans=(ans*a)%c;
            a=(a*a)%c;
            b=b>>1;
        }
        return ans;
    }
    int main()
    {
        cin>>a>>b>>k>>n>>m;
        for(int i=0;i<=1000;++i)
        {
            yh[i][0]=1;
            for(int j=1;j<=i;++j)
            yh[i][j]=(yh[i-1][j]+yh[i-1][j-1])%10007;
        }
        ans=(quickpow(a,n,10007)*quickpow(b,m,10007)*yh[k][m])%10007;
        cout<<ans;
        return 0;
    }
    
  • 2
    @ 2017-10-16 11:23:33

    根据二项式定理x^n*y^m的系数可表示为a^n*b^m*C(k,m)。
    所以本题可以用杨辉三角递推c[i][j]=c[i-1][j]+c[i-1][j-1].

    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    #include <cmath>
    #define LL long long 
    using namespace std;
    template <class _T> inline void read(_T &_x) {
        int _t; bool _flag=false;
        while((_t=getchar())!='-'&&(_t<'0'||_t>'9')) ;
        if(_t=='-') _flag=true,_t=getchar(); _x=_t-'0';
        while((_t=getchar())>='0'&&_t<='9') _x=_x*10+_t-'0';
        if(_flag) _x=-_x;
    }
    const int MOD=10007;
    LL a,b,k,n,m,C[1005][1005];
    
    void get(int x,int y){
        C[1][0]=C[1][1]=1;
        for(int i=2;i<=x;++i){
            C[i][0]=1;
            for(int j=1;j<=y;++j)
                C[i][j]=C[i-1][j-1]+C[i-1][j],C[i][j]%=MOD;
        }
    }
    LL ks(LL x,int y){
        LL s=1;
        while(y){
            if(y&1)s*=x,s%=MOD;
            y>>=1;
            x*=x;
            x%=MOD;
        }
        return s;
    }
    int main(){
        read(a),read(b),read(k),read(n),read(m);
        get(k,m);
        printf("%lld",((C[k][m]*ks(a,n)%MOD)*ks(b,m))%MOD);
        return 0;
    }
    
  • 1
    @ 2018-11-01 22:24:08

    每人用逆元法吗?
    那我来一个(复杂度nlogn)
    (虽然正解是用杨辉,但这个代码可以过k=100000的数据!(不过要把里面的数全都换成long long))

    //求a^n*b^m*C(k,n)
    //快速幂求前两个,组合数用扩欧求逆元进行除法计算
    //组合数有更好的求法就是杨辉三角形打表(但是复杂度较高,我这里给个更好的)
    //总复杂度O(nlogn) 
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<algorithm>
    #define mod 10007
    using namespace std;
    int n,m,k,e,f;
    typedef long long ll;
    int pow(int a,int x){//快速幂(注意要开long long!!!) 
        if(a==0) return 0;
        ll w=a,res=1,i=x;
        while(i){
            if(i&1) res=(res*w)%mod;
            w=(w*w)%mod;
            i>>=1;
        }
        return (int)res;
    }
    int exgcd(int a,int b,int &x,int &y){//扩欧 
        if(b==0){
            x=1;
            y=0;
            return a;
        }
        int res=exgcd(b,a%b,x,y);
        int tmp=x;
        x=y;
        y=tmp-(a/b)*y;
        return res;
    }
    int getC(int a,int b){//下标b,上标a ,求组合数 
        int i,res=1,x,y;
        if(a>b/2) a=b-a;
        for(i=b;i>b-a;i--){
            res=(res*i)%mod;
        }
        for(i=2;i<=a;i++){
            exgcd(i,mod,x,y);
            if(x<0) x=(x%mod)+mod;
            res=(res*x)%mod;
        }
        return res;
    }
    int main(){
        int ans=1;
        cin>>e>>f>>k>>n>>m;
        ans=(ans*pow(e,n))%mod;
        ans=(ans*pow(f,m))%mod;
        ans=(ans*getC(n,k))%mod;
        cout<<ans;
        return 0;
    }
    
  • 1
    @ 2017-10-27 16:53:58

    杨辉三角预处理
    O(k^2+lgn)

    #include <iostream>
    #include <iomanip>
    #include <cmath>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <algorithm>
    #include <cctype>
    #include <vector>
    #include <queue>
    using namespace std;
    int y[1001][1001];
    int a,b,k,n,m;
    void yhsj()
    {
        for(int i=0;i<=1000;i++)
        {
            y[i][0]=1;
            y[i][i]=1;
        }
        for(int i=2;i<=1000;i++)
        {
            for(int j=1;j<i;j++)
            {
                y[i][j]=(y[i-1][j-1]+y[i-1][j])%10007;
            }
        }
        return;
    }
    long long zym(int a,int b)
    {
        int ans=1;
        while(b)
        {
            if(b%2==1)
            {
                ans=(ans*a)%10007;
            }
            a=(a*a)%10007;
            b>>=1;
        }
        return ans%10007;
    }
    int main()
    {
        yhsj();
        scanf("%d%d%d%d%d",&a,&b,&k,&n,&m);
        a%=10007;
        b%=10007;
        printf("%d",(y[k][m]*zym(a,n)*zym(b,m))%10007);
        return 0;
    }
    
  • 1
    @ 2017-08-22 13:46:15
    #include <cmath>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <iomanip>
    #include <iostream>
    #include <algorithm>
    #include <vector>
    #include <deque>
    #include <set>
    #include <limits>
    #include <string>
    #include <sstream>
    using namespace std;
    
    const int key=10007,oo_min=0xcfcfcfcf,oo_max=0x3f3f3f3f;
    
    int a,b,t,n,m;
    int c[1000+1][1000+1];
    
    int main()
    {
        while (~scanf("%d%d%d%d%d",&a,&b,&t,&n,&m))
        {
            for (int i=0;i<=t;i++)
                c[i][0]=c[i][i]=1;
            for (int i=2;i<=t;i++)
                for (int j=1;j<i;j++)
                    c[i][j]=(c[i-1][j-1]+c[i-1][j])%key;
            a%=key,b%=key;
            int temp_1=1,temp_2=1;
            for (int i=1;i<=n;i++)
                temp_1=(temp_1*a)%key;
            for (int i=1;i<=m;i++)
                temp_2=(temp_2*b)%key;
            int ans=(((c[t][n]*temp_1)%key)*temp_2)%key;
            printf("%d\n",ans);
        }
    }
    
  • 1
    @ 2016-10-23 15:05:59

    #include<iostream>
    #include<cstdio>
    using namespace std;
    typedef long long ll;
    const int MOD=10007;
    int n,m,a,b,k,ans=1;
    int yh[1003][1010]={0};

    void build_yh()
    {
    yh[0][0]=yh[1][0]=yh[1][1]=1;
    for(int i=2;i<=k;i++)
    {
    yh[i][0]=yh[i][i]=1;
    for(int j=1;j<i;j++)
    yh[i][j]=(yh[i-1][j-1]+yh[i-1][j])%MOD;
    }
    }
    int main()
    {
    scanf("%d%d%d%d%d",&a,&b,&k,&n,&m);
    a%=MOD;b%=MOD;
    build_yh();
    for(int i=0;i<n;i++)
    ans=ans*a%MOD;
    for(int i=0;i<m;i++)
    ans=ans*b%MOD;
    ans=ans*yh[k][n]%MOD;

    cout<<ans;
    return 0;
    }

  • 1
    @ 2016-10-11 22:27:16

    这道题用杨辉三角计算C(k,m)!!!

    #include<iostream>
    #include<cstdio>
    using namespace std;
    int a,b,k,n,m;
    int f[1005][1005]={0};
    int pow(int p,int time){
    int i,re=1;
    p=p%10007;
    for(i=1;i<=time;i++){
    re=(re*p)%10007;
    }
    return re;
    }
    int main(){
    //freopen("factor9.in","r",stdin);
    int i,j,ans=1;
    cin>>a>>b>>k>>n>>m;
    int r1,r2,r3;
    r1=pow(a,n);
    r2=pow(b,m);
    f[1][1]=1;f[1][2]=1;
    for(i=2;i<=k;i++){
    for(j=1;j<=i+1;j++){
    f[i][j]=(f[i-1][j-1]+f[i-1][j])%10007;
    }
    }
    r3=f[k][m+1];
    ans=(((r1*r2)%10007)*r3)%10007;
    cout<<ans;
    return 0;
    }

  • 0
    @ 2018-09-09 18:41:19
    #include <cstdio>
    using namespace std;
    
    int pascal[1002][1002];
    
    int pwr(int base, int exponent)
    {
        int sqr[10];
        sqr[0] = base;
        for (int i = 1; i < 10; i++) {
            sqr[i] = sqr[i - 1] * sqr[i - 1];
            sqr[i] %= 10007;
        }
        int ans = 1;
        for (int i = 0; i < 10; i++) {
            if (exponent & (1 << i)) {
                ans *= sqr[i];
                ans %= 10007;
            }
        }
        return ans;
    }
    
    int main()
    {
        int a, b, k, n, m;
        scanf("%d%d%d%d%d", &a, &b, &k, &n, &m);
        pascal[1][1] = 1;
        for (int i = 2; i <= 1001; i++) {
            for (int j = 1; j <= 1001; j++) {
                pascal[i][j] = (pascal[i - 1][j] + pascal[i - 1][j - 1]) % 10007;
            }
        }
        printf("%d", (((pwr(a%10007, n) * pwr(b%10007, m)) % 10007) * pascal[k+1][n+1]) % 10007);
        return 0;
    }
    
    
  • 0
    @ 2017-10-14 21:04:21

    miao...

    #include <iostream>
    using namespace std;
    #define For(aHJEfaks, fwbjkWK, AENGIBv) for (int aHJEfaks = fwbjkWK; aHJEfaks <= AENGIBv; ++aHJEfaks)
    #define For2(aHJEfaks, fwbjkWK, AENGIBv) for (auto aHJEfaks = fwbjkWK; aHJEfaks != AENGIBv; ++aHJEfaks)
    long long a, b, k, n, m;
    long long PA[2000], PB[2000], C[2000][2000];
    long long pa(long long x){
        if (PA[x] != 0){
            return PA[x];
        }
        if (x == 0){
            return PA[x] = 1;
        }
        return PA[x] = (pa(x - 1) * a) % 10007;
    }
    long long pb(long long x){
        if (PB[x] != 0){
            return PB[x];
        }
        if (x == 0){
            return PB[x] = 1;
        }
        return PB[x] = (pb(x - 1) * b) % 10007;
    }
    long long c(long long x, long long y){
        if (x < y || y < 0){
            return 0;
        }
        if (y == 0){
            return 1;
        }
        if (x == y){
            return 1;
        }
        if (C[x][y] != 0){
            return C[x][y];
        }
        return C[x][y] = (c(x - 1, y - 1) + c(x - 1, y)) % 10007;
    }
    int main(){
        cin >> a >> b >> k >> n >> m;
        cout << (c(k, m) * pa(n) * pb(m)) % 10007 << endl;
        return 0;
    }
    
  • 0
    @ 2016-11-18 10:33:56

    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #define Q 10007

    int main(){
    long long q=1,a,b,f1[1200]={0},f2[1200]={0};
    int k,n,m;
    std::cin>>a>>b>>k>>n>>m;
    f1[0]=1,f1[1]=1;
    for(int i=2;i<=k;i++){
    memcpy(f2,f1,sizeof(f1));
    f1[0]=f2[0];
    for(int j=1;j<=i;j++)
    f1[j]=(f2[j-1]+f2[j])%Q;
    }
    q=f1[n];
    for(int i=1;i<=n;i++)
    q=(q*a)%Q;
    for(int i=1;i<=m;i++)
    q=(q*b)%Q;

    std::cout<<q;
    return 0;
    }

  • 0
    @ 2016-10-23 15:06:25

    要注意维护不要越界;

  • 0
    @ 2016-09-21 22:13:42

    想公式花了10分钟 敲高精度花了40分钟
    最后发现要MOD10007 发现不用高精度.....
    公式:i从1到k循环,j从1到k-1,每一次:
    result[j+1]+=a x q[j];
    result[j]+=b x q[j];
    q为上一步的结果 result为这一步的结果
    q[p]=m 表示(x^p) * (y^j-p)项的系数=m
    代码
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #define M 10007

    long long q[1010]={0};
    long long temp[1010]={0};

    int main(){
    int k,n,m;
    long long a,b;
    std::cin>>a>>b>>k>>n>>m;
    q[0]=b;
    q[1]=a;
    for(int i=2;i<=k;i++){
    memset(temp,0,sizeof(temp));
    for(int j=0;j<=k-1;j++){
    temp[j+1]+=a*q[j];
    temp[j]+=b*q[j];
    }
    for(int j=0;j<=k;j++)
    temp[j]%=M;
    memcpy(q,temp,sizeof(q));
    }
    std::cout<<q[n];
    return 0;
    }

  • 0
    @ 2016-09-14 18:47:28

    这题纯粹考高中数学
    ```c++
    #include<iostream>
    #include<cstdio>
    using namespace std;

    const int MOD = 10007;
    const int maxk = 1000 + 10;

    int a, b, k, n, m;
    int C[maxk][maxk];

    int main ()
    {
    // freopen("in.txt", "r", stdin);
    cin >> a >> b >> k >> n >> m;
    int ans = 1;
    a = a % MOD; b = b % MOD;
    for (int i = 0; i < n; i++)
    ans = ans * a % MOD;
    for (int i = 0; i < m; i++)
    ans = ans * b % MOD;

    for (int i = 0; i <= k; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++)
    C[i][j] = (C[i-1][j-1] + C[i-1][j]) % MOD;
    }

    ans = ans * C[k][n] % MOD;
    cout << ans << "\n";
    return 0;
    }
    ```

  • 0
    @ 2016-07-16 08:50:02

    评测状态 Accepted
    题目 P1739 计算系数
    递交时间 2016-07-16 08:49:21
    代码语言 C++
    评测机 ShadowShore
    消耗时间 15 ms
    消耗内存 556 KiB
    评测时间 2016-07-16 08:49:22
    评测结果

    编译成功

    foo.cpp: In function 'int main()':
    foo.cpp:54:48: warning: format '%d' expects argument of type 'int', but argument 2 has type 'long long int' [-Wformat=]
    printf("%d\n",C(k,n)%P*qpow(a,n)%P*qpow(b,m)%P);
    ^

    测试数据 #0: Accepted, time = 0 ms, mem = 556 KiB, score = 10

    测试数据 #1: Accepted, time = 0 ms, mem = 552 KiB, score = 10

    测试数据 #2: Accepted, time = 0 ms, mem = 552 KiB, score = 10

    测试数据 #3: Accepted, time = 0 ms, mem = 556 KiB, score = 10

    测试数据 #4: Accepted, time = 0 ms, mem = 556 KiB, score = 10

    测试数据 #5: Accepted, time = 0 ms, mem = 552 KiB, score = 10

    测试数据 #6: Accepted, time = 0 ms, mem = 552 KiB, score = 10

    测试数据 #7: Accepted, time = 15 ms, mem = 552 KiB, score = 10

    测试数据 #8: Accepted, time = 0 ms, mem = 556 KiB, score = 10

    测试数据 #9: Accepted, time = 0 ms, mem = 556 KiB, score = 10

    Accepted, time = 15 ms, mem = 556 KiB, score = 100
    代码

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<set>
    #include<ctime>
    #include<vector>
    #include<cmath>
    #include<algorithm>
    #define P 10007
    #define N 50005
    #define ll long long
    #define INF 1000000000
    using namespace std;
    int a,b,k,n,m;
    void exgcd(int a,int b,int &x,int &y)
    {
    if(!b){x=1;y=0;}
    else{exgcd(b,a%b,y,x);y-=x*(a/b);}
    }
    long long qpow(long long x,long long y)
    {
    int sum=1;
    while(y>0)
    {
    if(y&1)sum=sum*x%P;
    x=x*x%P;
    y>>=1;
    }
    return sum;
    }
    int ine(int T)
    {
    int x,y;
    exgcd(T,P,x,y);
    x%=P;
    while(x<=0)x+=P;
    return x;
    }
    int C(int n,int m)
    {
    int s1=1,s2=1;
    if(m>n-m)m=n-m;
    for(int i=1;i<=m;i++)
    {
    s1=s1*(n-i+1)%P;
    s2=s2*i%P;
    }
    return s1*ine(s2)%P;
    }
    int main()
    {
    scanf("%d%d%d%d%d",&a,&b,&k,&n,&m);
    printf("%d\n",C(k,n)%P*qpow(a,n)%P*qpow(b,m)%P);
    return 0;
    }

  • 0
    @ 2016-07-12 11:43:34

    二项式展开后发现就是求C(k,m)×(a^n)×(b^m),递推时间长了点但是能过。
    ~~~
    #include <cstdio>
    int a,b,n,m,ans,c[1005];
    int C(int u,int v){
    if(v>u-v) v=u-v;
    for(int i=0;i<=v;i++) c[i]=1;
    for(int i=1;i<=u-v;i++)
    for(int j=1;j<=v;j++)
    c[j]=(c[j]+c[j-1])%10007;
    return c[v];
    }
    int P(int u,int v){//power
    int x=1;
    for(int i=1;i<=v;i++)
    x=(x*(u%10007))%10007;
    return x;
    }
    int main(){
    scanf("%d%d%*d%d%d",&a,&b,&n,&m);
    ans=((C(n+m,m)*P(a,n))%10007*P(b,m))%10007;
    printf("%d\n",ans);
    return 0;
    }

  • 0
    @ 2015-11-07 22:36:29

    评测状态 Wrong Answer
    题目 P1739 计算系数
    递交时间 2015-11-07 22:35:48
    代码语言 C
    评测机 VijosEx
    消耗时间 60 ms
    消耗内存 4176 KiB
    评测时间 2015-11-07 22:35:49
    评测结果
    编译成功

    测试数据 #0: Accepted, time = 0 ms, mem = 4168 KiB, score = 10
    测试数据 #1: Accepted, time = 0 ms, mem = 4176 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 4168 KiB, score = 10
    测试数据 #3: Accepted, time = 15 ms, mem = 4168 KiB, score = 10
    测试数据 #4: WrongAnswer, time = 0 ms, mem = 4172 KiB, score = 0
    测试数据 #5: WrongAnswer, time = 15 ms, mem = 4176 KiB, score = 0
    测试数据 #6: WrongAnswer, time = 0 ms, mem = 4168 KiB, score = 0
    测试数据 #7: Accepted, time = 15 ms, mem = 4168 KiB, score = 10
    测试数据 #8: Accepted, time = 15 ms, mem = 4168 KiB, score = 10
    测试数据 #9: WrongAnswer, time = 0 ms, mem = 4168 KiB, score = 0
    WrongAnswer, time = 60 ms, mem = 4176 KiB, score = 60
    代码
    #include<stdio.h>
    #define MAX 1001
    int pangle[MAX][MAX+1];
    int power(int x,int k,int c)
    {
    int temp=1;
    x=x%c;
    while(k)
    {
    if(k%2==1)
    temp=(temp*x)%c;
    k=k/2;
    x=(x*x)%c;
    }
    return temp;
    }
    int main()
    {
    int ans=1,i,j,a,b,k,n,m;
    pangle[0][1]=1;
    for(i=1;i<MAX;i++)
    for(j=1;j<MAX+1;j++)
    pangle[i][j]=(pangle[i-1][j]+pangle[i-1][j-1])%10007;
    scanf("%d%d%d%d%d",&a,&b,&k,&n,&m);
    ans*=power(a,n,10007);
    ans*=power(b,m,10007);
    ans=ans*pangle[k][m+1]%10007;
    printf("%d",ans);
    return 0;
    }

    who can tell me why

  • 0
    @ 2015-10-26 08:34:05

    题目里给出的m没用吗?
    我不会 去搜了个代码 发现他根本没用到m
    但是AC了
    谁能告诉我这是毛线定理
    跪谢神犇

    var
    f:array[-1..1001,-1..1001] of int64;
    a,b,m,n,k:int64;
    i,j:longint;
    begin
    readln(a,b,k,n,m); m:=0;
    f[1,0]:=b;
    f[1,1]:=a;
    for i:=2 to k do
    for j:=0 to k do
    f[i,j]:=(f[i-1,j]*b+f[i-1,j-1]*a) mod 10007;
    writeln(f[k,n]);
    end.

    • @ 2016-07-12 11:34:15

      题目不是说了K=N+M么?

  • 0
    @ 2015-10-20 22:38:47

    错了好多次啊
    递推的
    记录信息
    评测状态 Accepted
    题目 P1739 计算系数
    递交时间 2015-10-20 22:37:06
    代码语言 C++
    评测机 VijosEx
    消耗时间 60 ms
    消耗内存 8292 KiB
    评测时间 2015-10-20 22:37:12
    评测结果
    编译成功

    foo.cpp: In function 'int main()':
    foo.cpp:29:19: warning: unknown conversion type character 'l' in format [-Wformat=]
    printf("%lld",ans);
    ^
    foo.cpp:29:19: warning: too many arguments for format [-Wformat-extra-args]
    测试数据 #0: Accepted, time = 0 ms, mem = 8260 KiB, score = 10
    测试数据 #1: Accepted, time = 0 ms, mem = 8260 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 8260 KiB, score = 10
    测试数据 #3: Accepted, time = 0 ms, mem = 8256 KiB, score = 10
    测试数据 #4: Accepted, time = 0 ms, mem = 8256 KiB, score = 10
    测试数据 #5: Accepted, time = 15 ms, mem = 8280 KiB, score = 10
    测试数据 #6: Accepted, time = 0 ms, mem = 8280 KiB, score = 10
    测试数据 #7: Accepted, time = 15 ms, mem = 8292 KiB, score = 10
    测试数据 #8: Accepted, time = 15 ms, mem = 8288 KiB, score = 10
    测试数据 #9: Accepted, time = 15 ms, mem = 8288 KiB, score = 10
    Accepted, time = 60 ms, mem = 8292 KiB, score = 100
    代码
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    using namespace std;
    const long long mod=10007;
    long long mem[1010][1010];
    long long work(int n,int k)
    {
    if(n==k)return 1;
    if(n==1)return k;
    if(mem[n][k])return mem[n][k];
    return mem[n][k]=(work(n,k-1)+work(n-1,k-1))%mod;
    }
    int main()
    {
    int a,b,k,n,m;
    scanf("%d%d%d%d%d",&a,&b,&k,&n,&m);
    long long ansx=1,ansy=1;
    a%=mod;//这两句话不加就会爆两个点
    b%=mod;
    for(int i=n,x=a;i;x=(x*x)%mod,i>>=1)
    if(i&1)ansx=(ansx*x)%mod;
    for(int i=m,y=b;i;y=(y*y)%mod,i>>=1)
    if(i&1)ansy=(ansy*y)%mod;
    long long ans=(ansx*ansy)%mod;

    ans=(ans*work(n,k))%mod;
    printf("%lld",ans);
    return 0;
    }

  • 0
    @ 2015-08-22 17:15:16

    #include<cstdio>
    #include<algorithm>
    #define mod 10007
    using namespace std;

    long long int num[2010][7000];

    int main()
    {
    int a, b, k, n, m;
    long long ans = 1;
    scanf("%d%d%d%d%d", &a, &b, &k, &n, &m);
    num[1][1] = num[1][2] = 1;
    for(int i=2; i<=k; i++)
    for(int j=1; j<=i+1; j++)
    num[i][j] = (num[i-1][j-1]+num[i-1][j])%mod;
    ans = num[k][m+1];
    for(int i=1; i<=n; i++)
    ans = (ans*a) % mod;
    for(int j=1; j<=m; j++)
    ans = (ans*b) % mod;
    printf("%lld", ans);
    return 0;
    }
    杨辉三角

    • @ 2015-08-22 17:15:36

      注意数组用long long

  • 0
    @ 2015-08-19 16:38:32

    const long long Mod=10007;
    cin>>a>>b>>k>>n>>m;
    c[0]=1;
    for (int i=1;i<=k;++i) {
    c[i]=c[i-1]*(k-i+1)/i;
    }
    c[m]%=Mod;
    cout<<(c[m]*qmod(a,k-m)*qmod(b,m))%Mod;

信息

ID
1739
难度
6
分类
数论 点击显示
标签
递交数
3814
已通过
1071
通过率
28%
被复制
2
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