A - Marjar Cola
签到题，x,y,a,b,都很小直接暴力。判断INF，只要判断次数有没有过多就行。
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
#define bug printf("*********\n");
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define mid (l+r)/2
#define chl 2*k+1
#define chr 2*k+2
#define lson l,mid,chl
#define rson mid,r,chr
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a));
const long long mod=1e9+7;
const int maxn=5e5+5;
const int INF=0x7fffffff;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
int main() {
int x,y,a,b;
int t;
scanf("%d",&t);
while(t--) {
int ans=0;
scanf("%d%d%d%d",&x,&y,&a,&b);
if(x==1||y==1)puts("INF");
else {
while(a>=x||y<=b) {
if(a>=x) {
a-=x;
a++;
b++;
ans++;
} else {
b-=y;
b++;
a++;
ans++;
}
// debug(a);
// debug(b);
if(ans>3e5){
break;
}
}
if(ans>3e5){
puts("INF");
}
else printf("%d\n",ans);
}
}
return 0;
}C - How Many Nines
打个表，比较考虑细节，真的难处理。
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
#define bug printf("*********\n");
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define mid (l+r)/2
#define chl 2*k+1
#define chr 2*k+2
#define lson l,mid,chl
#define rson mid,r,chr
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a));
const long long mod=1e9+7;
const int maxn=1e4+5;
const int INF=0x7fffffff;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
int Year[maxn], Mon[35], pre1[maxn], pre2[35];
int hh[13], mon[13];
bool leap(int y){
if(y%400==0)return 1;
if(y%100==0)return 0;
if(y%4==0)return 1;
return 0;
}
int nine(int y){
int cnt = 0;
while(y){
if(y%10==9)cnt++;
y/=10;
}
return cnt;
}
void init(){
Mon[1]=Mon[3]=Mon[5]=Mon[7]=Mon[8]=Mon[10]=Mon[12]=3;///31
Mon[2]=2;
Mon[4]=Mon[6]=Mon[9]=Mon[11] = 3;///30
Mon[9] += 30;
hh[1]=hh[3]=hh[5]=hh[7]=hh[8]=hh[10]=hh[12]=31;
hh[2] = 28;
hh[4]=hh[6]=hh[9]=hh[11]=30;
mon[0] = 0;
for(int i = 1; i <= 12; ++i){
mon[i] = hh[i]+mon[i-1];
}
int sum = 0;pre1[1999]=pre2[0] = 0;
for(int i = 1; i <= 12; ++i){
sum += Mon[i];
pre2[i] = pre2[i-1]+Mon[i];
}
for(int i = 2000; i <= 9999; ++i){
int tmp = nine(i);
if(tmp){
Year[i] = sum + tmp*365;
if(leap(i))Year[i] += tmp + 1;
}else{
Year[i] = sum;
if(leap(i))++Year[i];
}
pre1[i] = pre1[i-1]+Year[i];
}
}
int get(int d){
if(d<=9)return 3;
if(d<=19)return 2;
if(d<=29)return 1;
return 0;
}
int main(){
#ifndef ONLINE_JUDGE
//freopen("E://ADpan//in.in", "r", stdin);
//freopen("E://ADpan//out.out", "w", stdout);
#endif
init();
int tim;scanf("%d",&tim);
while(tim--){
int y1,m1,d1,y2,m2,d2;
scanf("%d%d%d%d%d%d",&y1,&m1,&d1,&y2,&m2,&d2);
if(y1==y2){
if(m1==m2){
int ans = get(d1)-get(d2+1);
if(m1==9)ans += d2-d1+1;
int tmp = nine(y1);
ans += tmp*(d2-d1+1);
printf("%d\n", ans);
continue;
}
int ans = pre2[m2-1]-pre2[m1];
if(leap(y1)&&2>m1&&2<m2)ans++;
ans += get(d1) + 3 - get(d2+1);
if(m1==2&&leap(y1)==false)ans--;
if(m1==9){
ans+=30-d1+1;
}
if(m2==9){
ans += d2;
}
int tmp = nine(y1);
if(tmp){
int day = mon[m2-1]-mon[m1] + (hh[m1] - d1 + 1) + d2;
if(leap(y1)&&2>m1&&2<m2)day++;
if(leap(y1)&&m1==2)day++;
ans += day*tmp;
}
printf("%d\n", ans);
}else{
int ans = pre1[y2-1]-pre1[y1];
int a = 0, b = 0, tmp = nine(y1),day;
a = pre2[12]-pre2[m1];
a += get(d1);
if(leap(y1)==0&&m1==2)a--;
if(leap(y1)&&m1==1)a++;
if(m1==9)a += 30-d1+1;
b = pre2[m2-1];
b += 3-get(d2+1);
if(leap(y2)&&m2>2)b++;
if(m2==9)b+=d2;
if(tmp){
day = mon[12]-mon[m1] + (hh[m1] - d1 + 1);
if(leap(y1)&&m1<=2)day++;
a += tmp * day;
}
tmp = nine(y2);
if(tmp){
day = mon[m2-1] + d2;
if(leap(y2)&&m2>2)day++;
b += tmp * day;
}
ans += a+b;
printf("%d\n", ans);
}
}
return 0;
}F - Knuth-Morris-Pratt Algorithm
签到题，KMP，判断一下次数就行。暴力判断也没事
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
#define bug printf("*********\n");
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define mid (l+r)/2
#define chl 2*k+1
#define chr 2*k+2
#define lson l,mid,chl
#define rson mid,r,chr
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a));
const long long mod=1e9+7;
const int maxn=5e5+5;
const int INF=0x7fffffff;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
int n;
char ch1[]="dog",ch2[]="cat";
int nex[2000];
void get_next(char *t,int lent){
nex[0] = -1;
for(int i = 0,k = -1;i < lent;){
if(k==-1||t[i] == t[k]){
++k;++i;
nex[i]=k;
}else k = nex[k];
}
}
int kmp(char *s,int lens,char *t,int lent){
if(lens<lent)return 0;
get_next(t,lent);
int i = 0, j = 0;
int cnt = 0;
while(i < lens&&j<lent) {
if(j==-1||s[i] == t[j]){
i++;j++;
if(j==lent){
cnt++;
j = nex[j];
}
}else j=nex[j];
}
return cnt;
}
int main() {
int t;
scanf("%d",&t);
while(t--){
char s[2000];
scanf("%s",s);
int l=strlen(s);
int ans=0;
ans+=kmp(s,l,ch1,3);
ans+=kmp(s,l,ch2,3);
printf("%d\n",ans);
}
return 0;
}G - Intervals
贪心，真的贪的内心崩溃。贪心策略，按L排序，每次判断3个区间，如果出现3个重合，丢去 R最大的，因为影响最大。
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
#define bug printf("*********\n");
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define mid (l+r)/2
#define chl 2*k+1
#define chr 2*k+2
#define lson l,mid,chl
#define rson mid,r,chr
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a));
const long long mod=1e9+7;
const int maxn=5e5+5;
const int INF=0x7fffffff;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
struct seg {
int st,en,id;
bool operator < (const seg a )const {
return en<a.en;
}
} sg[maxn];
bool cmp(seg a,seg b) {
if(a.st==b.st) {
return a.en<b.en;
}
return a.st<b.st;
}
int res[maxn];
int main() {
int n,t;
scanf("%d",&t);
while(t--) {
scanf("%d",&n);
for(int i=0; i<n; i++) {
scanf("%d%d",&sg[i].st,&sg[i].en);
sg[i].id=i+1;
}
sort(sg,sg+n,cmp);
priority_queue<seg>q;
int ans=0;
for(int i=0; i<n; i++) {
if(q.size()<2) {
q.push(sg[i]);
} else {
q.push(sg[i]);
seg s[3];
for(int i=0; i<3; i++) {
s[i]=q.top();
q.pop();
}
if(s[0].st<=s[2].en&&s[1].st<=s[2].en) {
res[ans++]=s[0].id;
q.push(s[1]);
q.push(s[2]);
} else {
q.push(s[0]);
q.push(s[1]);
}
}
}
sort(res,res+ans);
printf("%d\n",ans);
for(int i=0; i<ans; i++) {
printf("%d",res[i]);
if(i+1==ans)printf("\n");
else printf(" ");
}
if(ans==0)puts("");
}
return 0;
}H - Seven-Segment Display
题意 ：1-9九个数，分别可以用后面7个0 1表示，下面给你n个数，每个数后面跟有7个01串，你可以交换n个数任意两列。如果可以通过交换表示出来输出YES 否则NO。
例子 :
7 0101011 1 1101011
把第2列和第5列交换，变成
1 1001111 7 0001111
与1 7 的表示匹配，所以输出YES
题解：这题只有9个数，暴力啊匹配，n^2都不会超时。。。；我是把每一列的状态用一个10进制数保存，每次能够从原来的数里面找与之匹配的状态，如果找不到，就输出NO；
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
#define bug printf("*********\n");
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define mid (l+r)/2
#define chl 2*k+1
#define chr 2*k+2
#define lson l,mid,chl
#define rson mid,r,chr
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a));
const long long mod=1e9+7;
const int maxn=5e5+5;
const int INF=0x7fffffff;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
int mp[10];
int n;
int k[10],v[10];
int p[10];
map<int,int> m;
int main() {
mp[1]=1001111;
mp[2]=10010;
mp[3]=110;
mp[4]=1001100;
mp[5]=100100;
mp[6]=100000;
mp[7]=1111;
mp[8]=0;
mp[9]=100;
int t;
p[0]=1;
for(int i=1; i<10; i++) {
p[i]=p[i-1]*10;
}
scanf("%d",&t);
while(t--) {
scanf("%d",&n);
for(int i=0; i<n; i++) {
scanf("%d%d",&k[i],&v[i]);
}
m.clear();
for(int i=0; i<7; i++) {
int sum=0;
for(int j=0; j<n; j++) {
sum+=mp[k[j]]/p[i]%10*p[j];
}
m[sum]++;
}
int flag=1;
for(int i=0; i<7; i++) {
int sum=0;
for(int j=0; j<n; j++) {
sum+=v[j]/p[i]%10*p[j];
}
m[sum]--;
if(m[sum]<0) {
flag=0;
break;
}
}
puts(flag?"YES":"NO");
}
return 0;
}J - Course Selection System
傻逼题,一开始还在想怎么推公式，结果发现，CI的和只有5e4，枚举所用能够到的状态H 的和最大不久行了，直接转换成了 0 1 背包。。。。。
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
#define bug printf("*********\n");
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define mid (l+r)/2
#define chl 2*k+1
#define chr 2*k+2
#define lson l,mid,chl
#define rson mid,r,chr
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a));
const long long mod=1e9+7;
const int maxn=5e4+5;
const int INF=0x7fffffff;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
int n;
int a[maxn];
int x[maxn],y[maxn];
int main() {
int t;
scanf("%d",&t);
while(t--) {
scanf("%d",&n);
for(int i=0; i<n; i++) {
scanf("%d%d",&x[i],&y[i]);
}
mem(a,-1);
a[0]=0;
for(int i=0; i<n; i++) {
for(int j=5e4-y[i]; j>=0; j--) {
if(a[j]!=-1) {
a[j+y[i]]=max(a[j+y[i]],a[j]+x[i]);
}
}
}
ll ans=0;
for(ll i=1;i<=5e4;i++){
if(a[i]!=-1)
ans=max(ans,1LL*a[i]*a[i]-i*a[i]-i*i);
}
printf("%lld\n",ans);
}
return 0;
}