1 /*
 2 第一题比较容易，注意一些细节就可以了。
 3 比如删除前导0不能删没了等等。
 4 */
 5 #define N 1500
 6 #include<iostream>
 7 using namespace std;
 8 #include<cstdio>
 9 #include<cstring>
10 char a[N],b[N];
11 int a1[N],b1[N],len;
12 void input()
13 {
14     scanf("%s%s",a+1,b+1);
15     len=strlen(a+1);
16     for(int i=1;i<=len;++i)
17       a1[i]=a[i]-‘0‘;
18     for(int i=1;i<=len;++i)
19       b1[i]=b[i]-‘0‘;
20 }
21 int main()
22 {
23     freopen("number.in","r",stdin);
24     freopen("number.out","w",stdout);
25     input();
26     for(int i=1;i<=len;++i)
27     {
28         if(a1[i]>b1[i]) b1[i]=11;
29         if(a1[i]<b1[i]&&b1[i]!=11) a1[i]=11;
30     }
31     int ia=1,ib=1;
32     bool flag=false;
33     while(ia<len&&(a1[ia]==0||a1[ia]==11))
34     {
35          if(a1[ia]==0) flag=true;
36          ia++;
37     }
38     if(a1[ia]==11) ia++;
39     if(ia>len&&(!flag)) printf("BOOM");
40     else{
41         for(int i=ia;i<=len;++i)
42            if(a1[i]!=11) printf("%d",a1[i]);
43         if(ia>len&&flag) printf("0");
44     }
45     printf("\n");
46     flag=false;
47     while(ib<len&&(b1[ib]==0||b1[ib]==11))
48     {
49         if(b1[ia]==0) flag=true;
50         ib++;
51     }
52     if(b1[ib]==11) ib++;
53     if(ib>len&&(!flag)) printf("BOOM");
54     else{
55         for(int i=ib;i<=len;++i)
56           if(b1[i]!=11) printf("%d",b1[i]);
57         if(ib>len&&flag) printf("0");
58     }
59     fclose(stdin);
60     fclose(stdout);
61     return 0;
62 }

 1 #define N 405
 2 #include<iostream>
 3 using namespace std;
 4 #include<cstdio>
 5 #include<cstring>
 6 typedef long long ll;
 7 int n,sum[N];
 8 ll f1[N][N],f2[N][N];
 9 void input()
10 {
11     scanf("%d",&n);
12     int x;
13     for(int i=1;i<=n;++i)
14     {
15         scanf("%d",&x);
16         sum[i]=sum[i-1]+x;
17     }
18 }
19 void DP()
20 {
21     for(int i=1;i<=n;++i)
22       for(int j=1;j<=n;++j)
23          f1[i][j]=f2[i][j]=(1<<31)-1;
24     for(int i=1;i<=n-1;++i)
25       f2[i][i+1]=sum[i+1]-sum[i-1];
26     for(int i=1;i<=n;++i)
27       f1[i][i]=0;
28     for(int len=3;len<=n;++len)
29       for(int i=1;i+len-1<=n;++i)
30       {
31           int j=i+len-1;
32           for(int k=i;k<j;++k) f2[i][j]=min(f2[i][j],f1[i][k]+f1[k+1][j]+sum[j]-sum[i-1]);
33           for(int k=i;k<j;++k) f1[i][j]=min(f1[i][j],f2[i][k]+f1[k+1][j]+sum[j]-sum[k]);
34           /*注意这个推f1这个方程中，所加的代价是sum[j]-sum[k]，因为i---k-1,这段石子的花费，我们在推f2的时候已经加过了，所以不能重复加了*/
35       }
36 }
37 int main()
38 {
39     freopen("merge.in","r",stdin);
40     freopen("merge.out","w",stdout);
41     input();
42     DP();
43     cout<<f1[1][n]<<endl;
44     fclose(stdin);
45     fclose(stdout);
46     return 0;
47 }

 1 #include<iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 typedef long long ll;
 5 #define rep(i,a,n) for(int i=a;i<=n;i++)
 6 ll pp=1000000007;
 7 ll mo(ll a,ll pp){if(a>=0 && a<pp)return a;a%=pp;if(a<0)a+=pp;return a;}
 8 ll powmod(ll a,ll b,ll pp)
 9 {ll ans=1;
10 for(;b;b>>=1,a=mo(a*a,pp))
11 if(b&1)ans=mo(ans*a,pp);
12 return ans;}
13 ll gcd(ll a,ll b){return (!b)?a:gcd(b,a%b);}
14 ll read(){
15     ll ans=0;
16     char last=‘ ‘,ch=getchar();
17     while(ch<‘0‘ || ch>‘9‘)last=ch,ch=getchar();
18     while(ch>=‘0‘ && ch<=‘9‘)ans=ans*10+ch-‘0‘,ch=getchar();
19     if(last==‘-‘)ans=-ans;
20     return ans;
21 }
22 int n,m,p1[110000],p2[110000];
23 ll b[110000],inv[110000];
24 ll C(int n,int m){
25     if(n<m || n==0)return 0;
26     return  b[n]*inv[n-m]%pp*inv[m]%pp;
27 }
28 int main(){
29     freopen("problem.in","r",stdin);
30     freopen("problem.out","w",stdout);
31     n=read();m=read();
32     b[0]=1;
33     rep(i,1,n)b[i]=b[i-1]*i%pp;
34     rep(i,0,n)inv[i]=powmod(b[i],pp-2,pp);
35     
36     rep(i,1,m){
37         int x=read(),y=read();
38         if(x==y){
39             puts("0\n");
40             return 0;
41         }
42         p1[x]=1;
43         p2[y]=1;
44     }
45     ll ans=b[n-m];
46     //    cout<<ans<<endl;
47     int s=0;
48     rep(i,1,n)
49         if(!p1[i] && !p2[i])++s;
50     rep(i,1,s)
51         if(i&1)ans=mo(ans-C(s,i)*b[n-m-i],pp);
52         else ans=mo(ans+C(s,i)*b[n-m-i],pp);
53     cout<<ans<<endl;
54     return 0;
55 }