51nod oj 1256 1119 1013 1083 1014 1081<组合数+逆元+快速幂，DP，枚举，树状数组>

来源：互联网发布：js获取url网页内容编辑：程序博客网时间：2024/06/14 03:33

题目链接：1256

求逆元用拓展欧几里德--详情请看点击打开链接

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL long long LL extend_gcd(LL a,LL b,LL &x,LL &y){if (b==0){x=1;y=0;return a;}else{LL t=extend_gcd(b,a%b,y,x);y-=x*(a/b);return t;}}int main(){LL n,m,x,y;scanf("%lld%lld",&m,&n);extend_gcd(m,n,x,y);x=(x%n+n)%n;printf("%lld\n",x);}

题目链接：1119

题解与 LightOJ - 1067 数论<100000左右的组合数取模求法《逆元》> 一样

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL long longLL dis[2000100],ni[2000100];LL mod=1000000007;LL po(LL xx,LL k){    LL lp=1,hu=xx;    while (k)    {        if (k%2==1)            lp=(lp*hu)%mod;        hu=(hu*hu)%mod;        k/=2;    }    return lp;}int main(){    int n,m;    scanf("%d%d",&n,&m);    int da=n+m-2;    int xi=n-1;    dis[0]=ni[0]=1;    for (int i=1;i<=da;i++)    {        dis[i]=dis[i-1]*i%mod;    }    ni[xi]=po(dis[xi],mod-2);    ni[da-xi]=po(dis[da-xi],mod-2);    LL ans=dis[da]*ni[xi]%mod*ni[da-xi]%mod;    printf("%lld\n",ans);    return 0;}

题目链接：1013

快速幂+逆元

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL long longLL mod=1000000007;LL extend(LL a,LL b,LL &x,LL &y){    if (b==0)    {        x=1;y=0;        return a;    }    else    {        LL t=extend(b,a%b,y,x);        y-=x*(a/b);        return t;    }}LL pp(LL x,LL k){    LL lp=1,hu=x;    while (k)    {        if (k%2)        {            lp=lp*hu%mod;        }        hu=(hu*hu)%mod;        k/=2;    }    return lp;}int main(){    int n;scanf("%d",&n);    LL p=pp(3,n+1);    p-=1;    LL x,y,a=2;    extend(a,mod,x,y);    x=(x%mod+mod)%mod;    LL ans=p*x%mod;    printf("%lld\n",ans);    return 0;}

题目链接：1083

简单dp

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL long longLL map[600][600],dp[600][600];int main(){    int n;scanf("%d",&n);    memset(map,0,sizeof(map));    memset(dp,0,sizeof(dp));    for (int i=1;i<=n;i++)        for (int j=1;j<=n;j++)        scanf("%lld",&map[i][j]);    for (int i=1;i<=n;i++)        for (int j=1;j<=n;j++)        dp[i][j]=max(dp[i-1][j],dp[i][j-1])+map[i][j];    printf("%lld\n",dp[n][n]);    return 0;}

题目链接：1014

枚举----无语--

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;long long i,a,p;int main(){    scanf("%lld%lld",&p,&a);    bool fafe=true;    for (i=1;i<=p;i++)    {        if (i*i%p==a)        {            if (fafe)            {                printf("%lld",i);                fafe=false;            }            else                printf(" %lld",i);        }    }    if (fafe)        printf("No Solution\n");    else        printf("\n");    return 0;}

题目链接：1081

题解：树状数组求和

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define low(x) x&-x#define LL long longint n;LL shu[50050];void ADD(int ii,LL xx){for (ii;ii<=n;ii+=low(ii))shu[ii]+=xx;}LL QUERY(int ii){LL lp=0;for (ii;ii>0;ii-=low(ii))lp+=shu[ii];return lp;}int main(){scanf("%d",&n);LL a;memset(shu,0,sizeof(shu));for (int i=1;i<=n;i++){scanf("%lld",&a);ADD(i,a);}int m;scanf("%d",&m);int b,c;while (m--){scanf("%d%d",&b,&c);printf("%lld\n",QUERY(b+c-1)-QUERY(b-1));}return 0;}

0 0