A. Chunga-Changa
#include <bits/stdc++.h>
using namespace std;
long long x, y, z;
int main() {
long long ans =
0, cnt =
0;
scanf("%lld%lld%lld", &x, &y, &
z);
ans = (x + y) /
z;
long long p1 = x % z, p2 = y %
z;
if(z - p1 < z -
p2) {
if(p2 >= z - p1) cnt = z -
p1;
} else {
if(p1 >= z - p2) cnt = z -
p2;
}
printf("%lld %lld\n", ans, cnt);
return 0;
}
View Code
B. Split a Number
By Golden_miner, contest: Codeforces Round #
567 (Div.
2), problem: (B) Split a Number, Accepted, #
#include <bits/stdc++.h>
using namespace std;
int N;
string s, ans, temp;
string add(
string a,
string b) {
string c;
int len1 =
a.length();
int len2 =
b.length();
int len =
max(len1, len2);
for(
int i = len1; i < len; i ++
)
a =
"0" +
a;
for(
int i = len2; i < len; i ++
)
b=
"0" +
b;
int ok =
0;
for(
int i = len -
1; i >=
0; i --
) {
char temp = a[i] + b[i] -
'0' +
ok;
if(temp >
'9') {
ok =
1;
temp -=
10;
}
else ok =
0;
c = temp +
c;
}
if(ok) c =
"1" +
c;
return c;
}
string DelZero(
string t) {
string ans =
"";
int temp =
0;
for(
int i =
0; t[i]; i ++
) {
if(t[i] !=
'0') {
temp =
i;
break;
}
}
ans = t.substr(temp, t.length() -
temp);
return ans;
}
string Sum(
int x) {
string num1 = s.substr(
0, x);
string num2 = s.substr(x, N -
x);
string ans =
add(num1, num2);
return ans;
}
bool Bigger(
string &a,
string &
b) {
if(a.length() > b.length())
return true;
else if(a.length() < b.length())
return false;
else {
int len =
a.length();
for(
int i =
0; i < len; i ++
) {
if(a[i] == b[i])
continue;
if(a[i] > b[i])
return true;
return false;
}
}
}
int main() {
scanf("%d", &
N);
cin >>
s;
ans =
s;
int pos = (N -
1) /
2, p =
pos;
int cnt =
0;
while(p >=
0 && cnt <=
1) {
if(s[p +
1] !=
'0') {
temp = Sum(p +
1);
if(Bigger(ans, temp)) ans =
temp;
cnt ++
;
}
p --
;
}
p =
pos;
cnt =
0;
while(p < N -
1 && cnt <=
0) {
if(s[p +
1] !=
'0') {
temp = Sum(p +
1);
if(Bigger(ans, temp)) ans =
temp;
cnt ++
;
}
p ++
;
}
for(
int i =
0; i < ans.length();i ++
) {
printf("%c", ans[i]);
}
return 0;
}
View Code
C. Flag
#include <bits/stdc++.h>
using namespace std;
char mp[
1010][
1010];
int dp[
1010][
1010];
int N, M;
int main() {
scanf("%d%d", &N, &
M);
for(
int i =
1; i <= N; i ++
)
scanf("%s", mp[i] +
1);
memset(dp, 0,
sizeof(dp));
for(
int i = N; i >=
1; i --
) {
for(
int j =
1; j <= M; j ++
)
if(mp[i][j] == mp[i +
1][j])
dp[i][j] = dp[i +
1][j] +
1;
else dp[i][j] =
1;
}
long long ans =
0;
for(
int i =
1; i <= N; i ++
) {
for(
int j =
1, k =
0; j <= M; j ++
) {
int h =
dp[i][j];
if(i +
3 * h -
1 <= N && mp[i][j] != mp[i + h][j] && dp[i + h][j] ==
h
&& mp[i + h][j] != mp[i +
2 * h][j] && dp[i +
2 * h][j] >=
h) {
if(k && mp[i][j] == mp[i][j -
1] && dp[i][j -
1] ==
h
&& mp[i + h][j] == mp[i + h][j -
1] && dp[i + h][j -
1] ==
h
&& mp[i +
2 * h][j] == mp[i +
2 * h][j -
1] && dp[i +
2 * h][j -
1] >=
h)
++
k;
else k =
1;
} else k =
0;
ans +=
k;
}
}
printf("%lld\n", ans);
return 0;
}
View Code
D. Irrigation
#include <bits/stdc++.h>
using namespace std;
#define MAX 500010
#define LL long long
#define MOD 1000000007
struct node{
int id;
LL q;
};
bool cmp(node a,node b){
return a.q<
b.q;
}
node rr[MAX];
int ans[MAX];
int s[MAX*
5];
int a[MAX];
vector<
int>
e[MAX];
void update(
int l,
int r,
int tt,
int k){
if(l==
r){
s[tt]++
;
return ;
}
int mid=(l+r)/
2;
if(mid<
k)
update(mid+
1,r,tt*
2+
1,k);
else
update(l,mid,tt*
2,k);
s[tt]=s[tt*
2]+s[tt*
2+
1];
}
int query(
int l,
int r,
int tt,
int k){
if(l==
r)
return l;
int mid=(l+r)/
2;
if(s[tt*
2]<
k)
return query(mid+
1,r,tt*
2+
1,k-s[tt*
2]);
else
return query(l,mid,tt*
2,k);
}
int main(){
int n,m,q,x;
scanf("%d%d%d",&n,&m,&
q);
for(
int i=
1;i<=n;i++
){
scanf("%d",&
x);
a[x]++
;
}
LL st=n+
1;
int sz=
0;
for(
int i=
1;i<=m;i++
){
e[a[i]].push_back(i);
st=
min((LL)a[i],st);
}
sz=
e[st].size();
for(
int i=
0;i<e[st].size();i++
){
update(1,m,
1,e[st][i]);
}
LL y,pre=
n;
for(
int i=
1;i<=q;i++
){
scanf("%lld",&
y);
rr[i].id=
i;
rr[i].q=
y;
}
sort(rr+
1,rr+
1+
q,cmp);
for(
int i=
1;i<=q;i++
){
LL y=rr[i].q-
pre;
if(st>
n){
ans[rr[i].id]=(y-
1)%m+
1;
}
else{
while(y>sz&&st<=
n){
y-=
sz;
pre+=
sz;
st++
;
sz+=
e[st].size();
for(
int j=
0;j<e[st].size();j++
){
update(1,m,
1,e[st][j]);
}
}
if(st>
n){
ans[rr[i].id]=(y-
1)%m+
1;
}
else
ans[rr[i].id]=query(
1,m,
1,y);
}
}
printf("\n");
for(
int i=
1;i<=q;i++
)
printf("%d\n",ans[i]);
return 0;
}
View Code
一万年更新一次 最近的活终于要做完了 开开心心
十一会有大把的时间可以刷题 8
转载于:https://www.cnblogs.com/zlrrrr/p/11593285.html
