51nod水题集

1000 A + B

1 #include<iostream>
2 using namespace std;
3 int main(){
4     int a,b;
5     cin>>a>>b;
6     cout<<a+b;
7     return 0;
8 }
View Code

1005 大数加法

 1 #include<iostream>
 2 #include<cmath>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<algorithm>
 6 using namespace std;
 7 const int MAXN=10005;
 8 struct bign{
 9     int num,f[MAXN],fl;
10     bign(){num=0; memset(f,0,sizeof(f)); fl=1;}
11     void read(){
12         char last=' ',ch=getchar();
13         while (ch<'0' || ch>'9') last=ch,ch=getchar();
14         while (ch>='0' && ch<='9') f[++num]=ch-'0',ch=getchar();
15         if (last=='-') fl=-1;
16         for (int i=1;i<=(num/2);i++) swap(f[i],f[num-i+1]); 
17     }
18     void out(){
19         if (fl==-1) printf("-");
20         for (int i=num;i>=1;i--) printf("%d",f[i]);
21     }
22 }a,b;
23 bign operator +(bign a,bign b){
24     bign ans; int num_=max(a.num,b.num);
25     for (int i=num_;i>=1;i--)
26         ans.f[i]=a.f[i]*a.fl+b.f[i]*b.fl;
27     for (int i=1;i<=num_;i++)
28         if (abs(ans.f[i])>9) ans.f[i+1]+=ans.f[i]/10,ans.f[i]=ans.f[i]%10;
29     ans.num=num_+1;
30     while (ans.f[ans.num]==0 && ans.num>1) ans.num--;
31     if (ans.f[ans.num]<0) {
32         ans.fl=-1;
33         for (int i=1;i<=ans.num;i++) ans.f[i]=-ans.f[i];
34     }
35     for (int i=1;i<=ans.num;i++)
36             if (ans.f[i]<0) ans.f[i]+=10,ans.f[i+1]--;
37     while (ans.f[ans.num]==0 && ans.num>1) ans.num--;
38     return ans;
39 }
40 int main(){
41     a.read(),b.read();
42     a=a+b; a.out();
43     return 0;
44 }
View Code

1006 最长公共子序列Lcs

 1 #include<iostream>
 2 #include<cmath>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<algorithm>
 6 using namespace std;
 7 #define N 1005 
 8 char ch1[N],ch2[N],ans[N];
 9 int f[N][N],nu;
10 struct node{
11     int i,j;    
12 }t[N][N];
13 void solve(int i,int j){
14     if (i==0 || j==0) return ; 
15     if (t[i][j].i==i-1 && t[i][j].j==j-1) ans[++nu]=ch1[i];
16     solve(t[i][j].i,t[i][j].j); 
17 }
18 int main(){
19     scanf("%s%s",ch1+1,ch2+1);
20     int len1=strlen(ch1+1),len2=strlen(ch2+1);
21     for (int i=1;i<=len1;i++)
22         for (int j=1;j<=len2;j++){
23             if (f[i][j]<f[i][j-1]) f[i][j]=f[i][j-1],t[i][j].i=i,t[i][j].j=j-1;
24             if (f[i][j]<f[i-1][j]) f[i][j]=f[i-1][j],t[i][j].i=i-1,t[i][j].j=j;
25             if (ch1[i]==ch2[j] && f[i][j]<f[i-1][j-1]+1) f[i][j]=f[i-1][j-1]+1,t[i][j].i=i-1,t[i][j].j=j-1;
26         } 
27     solve(len1,len2);
28     for (int i=nu;i>=1;i--)
29         printf("%c",ans[i]);
30     return 0;
31 }
View Code

1008 N的阶乘 mod P

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm> 
 4 using namespace std;
 5 typedef long long ll;
 6 ll read(){
 7     char last=' ',ch=getchar(); ll ans=0;
 8     while (ch<'0'||ch>'9') last=ch,ch=getchar();
 9     while (ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
10     if (last=='-') ans=-ans; return ans;
11 }
12 int main(){
13     ll n=read(),p=read(),ans=1LL;
14     for (ll i=1;i<=n;i++) ans=(ans*i)%p;
15     printf("%lld",ans);
16     return 0;
17 }
View Code

1011 最大公约数GCD

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm> 
 4 using namespace std;
 5 typedef long long ll;
 6 ll read(){
 7     char last=' ',ch=getchar(); ll ans=0;
 8     while (ch<'0'||ch>'9') last=ch,ch=getchar();
 9     while (ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
10     if (last=='-') ans=-ans; return ans;
11 }
12 int gcd(int a,int b){
13     return (b==0)?a:gcd(b,a%b);
14 }
15 int main(){
16     int a=read(),b=read();
17     printf("%d",gcd(a,b));
18     return 0;
19 }
View Code

1012 最小公倍数LCM

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm> 
 4 using namespace std;
 5 typedef long long ll;
 6 ll read(){
 7     char last=' ',ch=getchar(); ll ans=0;
 8     while (ch<'0'||ch>'9') last=ch,ch=getchar();
 9     while (ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
10     if (last=='-') ans=-ans; return ans;
11 }
12 int gcd(int a,int b){
13     return (b==0)?a:gcd(b,a%b);
14 }
15 int main(){
16     int a=read(),b=read();
17     printf("%lld",(ll)a*(ll)b/(ll)gcd(a,b));
18     return 0;
19 }
View Code

 1018 排序

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm> 
 4 using namespace std;
 5 typedef long long ll;
 6 ll read(){
 7     char last=' ',ch=getchar(); ll ans=0;
 8     while (ch<'0'||ch>'9') last=ch,ch=getchar();
 9     while (ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
10     if (last=='-') ans=-ans; return ans;
11 }
12 int a[50005];
13 int main(){
14     int n=read();
15     for (int i=1;i<=n;i++) a[i]=read();
16     sort(a+1,a+n+1);
17     for (int i=1;i<=n;i++) printf("%d\n",a[i]);
18     return 0;
19 }
View Code

1019 逆序数

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm> 
 4 using namespace std;
 5 typedef long long ll;
 6 ll read(){
 7     char last=' ',ch=getchar(); ll ans=0;
 8     while (ch<'0'||ch>'9') last=ch,ch=getchar();
 9     while (ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
10     if (last=='-') ans=-ans; return ans;
11 }
12 int a[50005],b[50005];
13 ll ans;
14 void merge(int l,int r){
15     int mid=(l+r)>>1,i=l,j=mid+1,nu=0;
16     while (i<=mid && j<=r){
17         while (i<=mid && a[i]<=a[j]) b[++nu]=a[i],i++;
18         while (j<=r && a[j]<a[i]) b[++nu]=a[j],j++,ans+=(ll)(mid-i+1);
19     }
20     while (i<=mid) b[++nu]=a[i],i++;
21     while (j<=r) b[++nu]=a[j],j++;
22     for (int i=1;i<=nu;i++) a[l+i-1]=b[i];
23 }
24 void msort(int l,int r){
25     if (l==r) return ;
26     int mid=(l+r)>>1; 
27     msort(l,mid);
28     msort(mid+1,r);
29     merge(l,r);
30 }
31 int main(){
32     int n=read();
33     for (int i=1;i<=n;i++) a[i]=read();
34     msort(1,n);
35     printf("%d",ans);
36     return 0;
37 }
View Code

1027 大数乘法

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm> 
 6 using namespace std;
 7 const int MAXN=2005;
 8 struct bign{
 9     int num,f[MAXN],fl;
10     bign(){num=0; memset(f,0,sizeof(f)); fl=1;}
11     void read(){
12         char last=' ',ch=getchar();
13         while (ch<'0' || ch>'9') last=ch,ch=getchar();
14         while (ch>='0' && ch<='9') f[++num]=ch-'0',ch=getchar();
15         if (last=='-') fl=-1;
16         for (int i=1;i<=(num/2);i++) swap(f[i],f[num-i+1]); 
17     }
18     void out(){
19         if (fl==-1) printf("-");
20         for (int i=num;i>=1;i--) printf("%d",f[i]);
21     }
22 }a,b;
23 bign operator *(bign a,bign b){
24     bign c; c.fl=a.fl*b.fl;
25     for (int i=1;i<=a.num;i++)
26         for (int j=1;j<=b.num;j++){
27             c.f[i+j-1]+=a.f[i]*b.f[j];
28         }
29     c.num=a.num+b.num;
30     for (int i=1;i<=c.num;i++) c.f[i+1]+=c.f[i]/10,c.f[i]%=10;
31     while (c.f[c.num]==0) c.num--;
32     return c;
33 }
34 int main(){
35     a.read(),b.read();
36     a=a*b; a.out();
37     return 0;
38 }
View Code

1046 A^B Mod C

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm> 
 4 using namespace std;
 5 typedef long long ll;
 6 ll read(){
 7     ll ans=0; char last=' ',ch=getchar();
 8     while (ch<'0'||ch>'9') last=ch,ch=getchar();
 9     while (ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
10     if (last=='-') ans=-ans; return ans;
11 }
12 int main(){
13     ll a=read(),b=read(),c=read(),ans=1;
14     for (;b;b>>=1,a=(a*a)%c) if (b&1) ans=(ans*a)%c;
15     printf("%lld",ans);
16     return 0;
17 }
View Code

1049 最大子段和

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm> 
 4 using namespace std;
 5 typedef long long ll;
 6 const int inf=1<<29;
 7 ll read(){
 8     ll ans=0; char last=' ',ch=getchar();
 9     while (ch<'0'||ch>'9') last=ch,ch=getchar();
10     while (ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
11     if (last=='-') ans=-ans; return ans;
12 }
13 ll f[50005],a[50005];
14 int main(){
15     int n=read(); ll ans=0; f[0]=-inf;
16     for (int i=1;i<=n;i++) a[i]=read();
17     for (int i=1;i<=n;i++) f[i]=max(f[i-1]+a[i],a[i]);
18     for (int i=1;i<=n;i++) ans=max(ans,f[i]);
19     printf("%lld",ans);
20     return 0;
21 }
View Code

1057 N的阶乘

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm> 
 6 using namespace std;
 7 typedef long long ll;
 8 const int MAXN=1000005;
 9 ll ans[MAXN];
10 int main(){
11     int n; scanf("%d",&n); ans[1]=1; int num=1;
12     for (int i=2;i<=n;i++) {
13         int c=0;
14         for(int j=1;j<=num;j++)
15         {
16             int tmp=i*ans[j]+c;
17             c=tmp/10000; ans[j]=tmp%10000;
18         }
19         while(c>0)
20         {
21             ans[++num]=c%10000; c/=10000;
22         }
23     }
24     printf("%d",ans[num]);
25     for (int i=num-1;i>=1;i--) printf("%04d",ans[i]);
26     return 0;
27 }
View Code

1058 N的阶乘的长度

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm> 
 6 using namespace std;
 7 typedef long long ll;
 8 ll read(){
 9     ll ans=0; char last=' ',ch=getchar();
10     while (ch<'0'||ch>'9') last=ch,ch=getchar();
11     while (ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
12     if (last=='-') ans=-ans; return ans;
13 }
14 int main(){
15     int n=read(); double ans=1;
16     for (int i=1;i<=n;i++) ans+=log10(i);
17     printf("%d",(int)ans);
18     return 0;
19 }
View Code

1066 Bash游戏

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 using namespace std;
 5 int read(){
 6     int ans=0; char last=' ',ch=getchar();
 7     while (ch<'0'||ch>'9') last=ch,ch=getchar();
 8     while (ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
 9     if (last=='-') ans=-ans; return ans;
10 }
11 int main(){
12     int T=read();
13     while (T--){
14         int n=read(),k=read();
15         if (n%(k+1)==0) printf("B\n");
16         else printf("A\n");    
17     }
18     return 0;
19 }
View Code

1069 Nim游戏2333

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 using namespace std;
 5 int read(){
 6     int ans=0; char last=' ',ch=getchar();
 7     while (ch<'0'||ch>'9') last=ch,ch=getchar();
 8     while (ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
 9     if (last=='-') ans=-ans; return ans;
10 }
11 int main(){
12     int n=read(),x=0;
13     for (int i=1;i<=n;i++) {
14         int a=read();
15         x^=a;
16     }
17     if (x==0) printf("B\n");
18     else printf("A\n");    
19     return 0;
20 }
View Code

1072 威佐夫游戏

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 using namespace std;
 5 int read(){
 6     int ans=0; char last=' ',ch=getchar();
 7     while (ch<'0'||ch>'9') last=ch,ch=getchar();
 8     while (ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
 9     if (last=='-') ans=-ans; return ans;
10 }
11 #define N 2000000
12 int f[N+5],nu;
13 int main(){
14     for (int i=1;i<=N;i++)
15     if (!f[i]){
16         if (i+nu+1>N) break;
17         f[i]=i+(++nu); f[i+nu]=i;
18     }
19     int T=read();
20     while (T--){
21         int a=read(),b=read();
22         if (f[a]==b) printf("B\n");
23         else printf("A\n");    
24     }
25     return 0;
26 }
View Code

1089 最长回文子串 V2（Manacher算法）2333

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 using namespace std;
 7 typedef long long ll;
 8 ll read(){
 9     ll ans=0; char last=' ',ch=getchar();
10     while (ch<'0'||ch>'9') last=ch,ch=getchar();
11     while (ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
12     if (last=='-') ans=-ans; return ans;
13 }
14 char ch[100005],s[200005];
15 int p[200005];
16 int main(){
17     scanf("%s",ch);
18     int n=strlen(ch),ans=0,Mi=0;
19     for (int i=0;i<=n;i++){
20         s[i+i+2]=ch[i];
21         s[i+i+1]='#';
22     }
23     s[0]='*';
24     for (int i=1;i<=n+n;i++){
25         if (Mi+p[Mi]>i) p[i]=min(p[Mi+Mi-i],Mi+p[Mi]-i);
26         else p[i]=1;
27         while (s[i-p[i]]==s[i+p[i]]) ++p[i];
28         if (Mi+p[Mi]<i+p[i]) Mi=i;
29         if (p[i]>ans) ans=p[i];
30     }
31     printf("%d",ans-1);
32     return 0;
33 }
View Code

1106 质数检测

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 using namespace std;
 7 typedef long long ll;
 8 ll read(){
 9     ll ans=0; char last=' ',ch=getchar();
10     while (ch<'0'||ch>'9') last=ch,ch=getchar();
11     while (ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
12     if (last=='-') ans=-ans; return ans;
13 }
14 int main(){
15     int T=read();
16     while (T--){
17         int n=read(),sn=sqrt(n),flag=0;
18         for (int i=2;i<=sn;i++)
19         if (n%i==0) {
20             flag=1;
21             break;
22         }
23         if (!flag) puts("Yes");
24         else puts("No");
25     }
26     return 0;
27 }
View Code

1118 机器人走方格

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 using namespace std;
 7 typedef long long ll;
 8 ll read(){
 9     ll ans=0; char last=' ',ch=getchar();
10     while (ch<'0'||ch>'9') last=ch,ch=getchar();
11     while (ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
12     if (last=='-') ans=-ans; return ans;
13 }
14 const int N=1005;
15 ll dp[N][N];
16 int main(){
17     int n=read(),m=read();
18     dp[1][1]=1;
19     for (int i=1;i<=n;i++)
20         for (int j=1;j<=m;j++)
21             if (!dp[i][j]) dp[i][j]=(dp[i-1][j]+dp[i][j-1])%1000000007; 
22     printf("%lld",dp[n][m]);
23 }
View Code

 

posted @ 2017-09-25 10:55  SXia  阅读(137)  评论(0)  编辑  收藏  举报