2017中国大学生程序设计竞赛 - 女生专场（重现）

#include<bits/stdc++.h>
using namespace std;

struct node
{
int flag,sum;
}a[10000];

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(a,0,sizeof(a));
int n,m;
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++)
{
int p,h,m;
char s[10];
scanf("%d%d:%d %s",&p,&h,&m,s);
if(a[p].flag==0)
{
if(strcmp(s,"AC")==0)
{
a[p].flag=1;
a[p].sum+=h*60+m;
}
else
{
a[p].sum+=20;
}
}
}
int sum=0,ans=0;
for(int i=1001;i<1001+n;i++)
{
if(a[i].flag==1)
{
sum++;
ans+=a[i].sum;
}
}
printf("%d %d\n",sum,ans);
}
}

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define met(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
typedef long long ll;
const int maxn = 3000+10;
ll dp[maxn];
struct node
{
ll x;
ll price;
}num[maxn];
ll dis[maxn];
int cmp(node a,node b)
{
return a.x<b.x;
}

int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
met(dis,0);
for(ll i=0;i<n;i++)
scanf("%lld%lld",&num[i].x,&num[i].price);
sort(num,num+n,cmp);
for(int i=0;i<n;i++)
{
if(i==0)
dis[i]=num[i].x;
else
dis[i]=dis[i-1]+num[i].x;
}
met(dp,inf);
dp[0]=num[0].price;
for(ll i=1;i<n;i++)
{
dp[i]=dp[0]+dis[i]-dis[0]*(i+1);
for(ll j=1;j<i;j++)
{
if(dp[i]>dp[j-1]+num[j].price+dis[i]-dis[j]-num[j].x*(i-j))
dp[i]=dp[j-1]+num[j].price+dis[i]-dis[j]-num[j].x*(i-j);
}
if(dp[i]>dp[i-1]+num[i].price)
dp[i]=dp[i-1]+num[i].price;
}
printf("%lld\n",dp[n-1]);
}
}

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define  met(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
const int maxn = 3000+10;
typedef long long ll;
ll dp[maxn][2];
struct node
{
ll x,price;
}num[maxn];

ll cmp(node a,node b)
{
return a.x<b.x;
}

int main()
{
ll n;
while(scanf("%lld",&n)!=EOF)
{
for(ll i=1;i<=n;i++)
scanf("%lld%lld",&num[i].x,&num[i].price);
sort(num+1,num+1+n,cmp);
met(dp,inf);
dp[0][0]=0;
dp[0][1]=0;
for(ll i=1;i<=n;i++)
{
dp[i][1]=min(dp[i-1][0],dp[i-1][1])+num[i].price;
ll ans=0;
for(ll j=i-1;j>=1;j--)
{
ans+=(num[j+1].x-num[j].x)*(i-j);
dp[i][0]=min(dp[i][0],dp[j][1]+ans);
}
}

4000
printf("%lld\n",min(dp
[1],dp
[0]));
}
}

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+10000;
int l
,r
;
int a
;
int main() {
int t;
scanf("%d",&t);
while(t--) {
int n;
scanf("%d",&n);
for(int i=1; i<=n; i++)
scanf("%d",&a[i]);
l[0]=0;
for(int i=1; i<=n; i++)
l[i]=__gcd(l[i-1],a[i]);
r[n+1]=0;
for(int i=n; i>=1; i--)
r[i]=__gcd(r[i+1],a[i]);
int ans=max(r[2],l[n-1]);
for(int i=2; i<=n-1; i++)
ans =max(__gcd(l[i-1],r[i+1]),ans);
printf("%d\n",ans);
}
return 0;
}

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll mod=1e9+7;
int main() {
int t;
scanf("%d",&t);
while(t--) {
ll n,k;
cin>>n>>k;
ll ans=0;
for(int i=1; i<=n; i++) {
if(k==0) {
ans+=1;
continue;
}
ll res=i;
for(int j=2; j<=k; j++) {
res%=mod;
res*=i;
res=res%mod;
}
ans%=mod;
ans+=res;
ans%=mod;
}
cout<<ans<<endl;
}
return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d",&t);
while(t--) {
int n,a[100005];
int flag=1;
scanf("%d",&n);
for(int i=2; i<=n; i++) {
scanf("%d",&a[i]);
}
int ans=0;
a[1]=2;
for(int i=n; i>=1; i--) {
if(a[i]==1)
ans++;
else
ans--;
if(ans<0)
flag=0;
}
if(flag&&n%2==0)
printf("Yes\n");
else
printf("No\n");
}
}

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD=1e9+7;
struct M {
ll p[4][4];
};
M mul(M AA,M BB) {
M aa;
int i,j,k;
memset(aa.p,0,sizeof(aa.p));
for(i=0; i<4; i++) {
for(j=0; j<4; j++) {
aa.p[i][j]=0;
for(k=0; k<4; k++) {
aa.p[i][j]+=((AA.p[i][k]*BB.p[k][j])%MOD);
}
aa.p[i][j]=aa.p[i][j]%MOD;
}
}
return aa;
}
M kuaisumi(M A,ll n) {
M res;
memset(res.p,0,sizeof(res.p));
for(int i=0; i<4; i++)
res.p[i][i]=1;
while(n) {
if(n&1)
res=mul(res,A);
A=mul(A,A);
n>>=1;
}
return res;
}
int main() {
M A= {1,1,0,0,0,0,1,0,1,0,0,1,0,0,0,0};
M B= {9,6,4,3,0,0,0,0,0,0,0,0,0,0,0,0};
ll n;
int t;
scanf("%d",&t);
while(t--) {

scanf("%lld",&n);
if(n<=5) {
if(n==2) {
printf("3\n");
continue;
}
if(n==3) {
printf("4\n");
continue;
}
if(n==4) {
printf("6\n");
continue;
}
if(n==5) {
printf("9\n");
continue;
}
}
M res=kuaisumi(A,n-5);
M ff=mul(B,res);
printf("%lld\n",ff.p[0][0]);
}
return 0;
}