2019江西省省赛

 

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 typedef long long ll;
 5 char c[1000];
 6 ll l,k1,k2,k3,k4,k,kk,d;
 7 
 8 int main(){
 9     while (~scanf("%lld",&l)) {
10         scanf("%s", c);
11         k1=k2=k3=k4=0;
12         for (int i = 0; i < l; i++) {
13             if (c[i] == 'a') k1++;
14             if (c[i] == 'v') k2++;
15             if (c[i] == 'i') k3++;
16             if (c[i] == 'n') k4++;
17         }
18         k = k2 * k1 * k3 * k4;
19         kk = l * l * l * l;
20         if (k == 0) {
21             printf("0/1\n");
22             continue;
23         }
24         d = __gcd(k, kk);
25         k = k / d;
26         kk = kk / d;
27         printf("%lld/%lld\n", k, kk);
28     }
29 }

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 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int a[2000],b[2000],c[2000],d[2000],t,f,n,m;
 5 int main() {
 6     while (~scanf("%d%d",&n,&m)) {
 7         for (int i = 1; i <= n; i++) {
 8             scanf("%d", &a[i]);
 9         }
10         for (int i = 1; i <= m; i++) {
11             scanf("%d", &b[i]);
12         }
13         t=0;
14         while (1) {
15             memset(c,0, sizeof(c));
16             memset(d,0, sizeof(d));
17             for (int i = 1; i <= n; i++) {
18                 c[a[i]] = 1;
19             }
20             for (int i = 1; i <= m; i++) {
21                 d[b[i] + t] = 1;
22             }
23             f=0;
24             for (int i=1;i<=1000;i++){
25                 if (c[i]&&d[i]==1){
26                     f=1;
27                     break;
28                 }
29             }
30             if (!f){
31                 printf("%d\n",t);
32                 break;
33             }
34             t++;
35         }
36     }
37 }

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 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int mod = 1000000007;
 4 long long quickpow(long long a, long long b) {
 5     if (b < 0) return 0;
 6     long long ret = 1;
 7     a %= mod;
 8     while(b) {
 9         if (b & 1) ret = (ret * a) % mod;
10         b >>= 1;
11         a = (a * a) % mod;
12     }
13     return ret;
14 }
15 long long inv(long long a) {
16     return quickpow(a, mod - 2);
17 }
18 int main() {
19     long long  n;
20     scanf("%lld",&n);
21     long long ans=((n+1)*inv(2*n))%mod;
22     printf("%lld\n",ans);
23     return 0;
24 }

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 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 typedef long long ll;
 5 ll n,b[1000],a[2000],m,lcm,x[2000],k,sum;
 6 
 7 ll gcd(ll a,ll b){
 8     if (!b){
 9         return a;
10     }
11     return gcd(b,a%b);
12 }
13 int main() {
14     while (~scanf("%lld%lld", &n,&m)) {
15         for (int i=1;i<=n;i++){
16             scanf("%lld",&a[i]);
17             b[a[i]]++;
18         }
19         lcm=1;
20         for (int i=1;i<=10;i++){
21             if (b[i]){
22                 lcm=i/gcd(i,lcm)*lcm;
23             }
24         }
25         for (int i=1;i<=n;i++){
26             x[i]=lcm/a[i];
27             sum+=x[i];
28         }
29         if (m%sum==0){
30             k=m/sum;
31             printf("Yes\n");
32             for (int i=1;i<=n;i++){
33                 printf("%lld",x[i]*k);
34                 if (i==n){
35                     printf("\n");
36                 }else{
37                     printf(" ");
38                 }
39             }
40         }else{
41             printf("No\n");
42         }
43     }
44 }

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1 #include<bits/stdc++.h>
2 using namespace std;
3 
4 int main()
5 {
6     int x,y;
7     scanf("%d%d",&x,&y);
8     printf("%d\n",(x*x-y*y)/4);
9 }

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