hdu 4902 Nice boat(2014多校訓練第4場1006)(暴力表記)
4803 ワード
Nice boat
Time Limit: 30000/15000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Problem Description
There is an old country and the king fell in love with a devil. The devil always asks the king to do some crazy things. Although the king used to be wise and beloved by his people. Now he is just like a boy in love and can’t refuse any request from the devil. Also, this devil is looking like a very cute Loli.
Let us continue our story, z*p(actually you) defeat the 'MengMengDa' party's leader, and the 'MengMengDa' party dissolved. z*p becomes the most famous guy among the princess's knight party.
One day, the people in the party find that z*p has died. As what he has done in the past, people just say 'Oh, what a nice boat' and don't care about why he died.
Since then, many people died but no one knows why and everyone is fine about that. Meanwhile, the devil sends her knight to challenge you with Algorithm contest.
There is a hard data structure problem in the contest:
There are n numbers a_1,a_2,...,a_n on a line, everytime you can change every number in a segment [l,r] into a number x(type 1), or change every number a_i in a segment [l,r] which is bigger than x to gcd(a_i,x) (type 2).
You should output the final sequence.
Input
The first line contains an integer T, denoting the number of the test cases.
For each test case, the first line contains a integers n.
The next line contains n integers a_1,a_2,...,a_n separated by a single space.
The next line contains an integer Q, denoting the number of the operations.
The next Q line contains 4 integers t,l,r,x. t denotes the operation type.
T<=2,n,Q<=100000 a_i,x >=0 a_i,x is in the range of int32(C++)
Output
For each test case, output a line with n integers separated by a single space representing the final sequence.
Please output a single more space after end of the sequence
Sample Input
Sample Output
題意:n個の数を与えて、それからこのn個の数に対して2種類の操作を行います:
1 l r xであれば、[l,r]区間の各数をxに変える.
もしそうなら 2 l r xは、[l,r]区間の数a_を比較するiとxのサイズ、a_の場合i>x,a_iがa_になるiとxの最大公約数.
最後にこのn個の数の最終的な値を出力します.
大神たちが線分樹を使っていたという問題を見ていますが、私のような弱い料理は線分樹を書くことができず、暴力で頼むしかありません.
解析:問題に従って直接シミュレーションして解くと、必ずタイムアウトします.したがって,逆方向に解くと,すなわち,各数に対して,後の操作のみが最終結果に影響し,最後の1操作前の操作はいずれも最終結果に影響しない.これにより、各数の最終結果を求める場合、スタックでこの数に影響する操作を格納することができ、1操作に遭遇するまで、または1操作に遭遇しなかった場合、初期値は入力されたa_である.i.それから問題の意味に従ってシミュレーションして解くといいです.
Time Limit: 30000/15000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Problem Description
There is an old country and the king fell in love with a devil. The devil always asks the king to do some crazy things. Although the king used to be wise and beloved by his people. Now he is just like a boy in love and can’t refuse any request from the devil. Also, this devil is looking like a very cute Loli.
Let us continue our story, z*p(actually you) defeat the 'MengMengDa' party's leader, and the 'MengMengDa' party dissolved. z*p becomes the most famous guy among the princess's knight party.
One day, the people in the party find that z*p has died. As what he has done in the past, people just say 'Oh, what a nice boat' and don't care about why he died.
Since then, many people died but no one knows why and everyone is fine about that. Meanwhile, the devil sends her knight to challenge you with Algorithm contest.
There is a hard data structure problem in the contest:
There are n numbers a_1,a_2,...,a_n on a line, everytime you can change every number in a segment [l,r] into a number x(type 1), or change every number a_i in a segment [l,r] which is bigger than x to gcd(a_i,x) (type 2).
You should output the final sequence.
Input
The first line contains an integer T, denoting the number of the test cases.
For each test case, the first line contains a integers n.
The next line contains n integers a_1,a_2,...,a_n separated by a single space.
The next line contains an integer Q, denoting the number of the operations.
The next Q line contains 4 integers t,l,r,x. t denotes the operation type.
T<=2,n,Q<=100000 a_i,x >=0 a_i,x is in the range of int32(C++)
Output
For each test case, output a line with n integers separated by a single space representing the final sequence.
Please output a single more space after end of the sequence
Sample Input
1
10
16807 282475249 1622650073 984943658 1144108930 470211272 101027544 1457850878 1458777923 2007237709
10
1 3 6 74243042
2 4 8 16531729
1 3 4 1474833169
2 1 8 1131570933
2 7 9 1505795335
2 3 7 101929267
1 4 10 1624379149
2 2 8 2110010672
2 6 7 156091745
1 2 5 937186357
Sample Output
16807 937186357 937186357 937186357 937186357 1 1 1624379149 1624379149 1624379149
題意:n個の数を与えて、それからこのn個の数に対して2種類の操作を行います:
1 l r xであれば、[l,r]区間の各数をxに変える.
もしそうなら 2 l r xは、[l,r]区間の数a_を比較するiとxのサイズ、a_の場合i>x,a_iがa_になるiとxの最大公約数.
最後にこのn個の数の最終的な値を出力します.
大神たちが線分樹を使っていたという問題を見ていますが、私のような弱い料理は線分樹を書くことができず、暴力で頼むしかありません.
解析:問題に従って直接シミュレーションして解くと、必ずタイムアウトします.したがって,逆方向に解くと,すなわち,各数に対して,後の操作のみが最終結果に影響し,最後の1操作前の操作はいずれも最終結果に影響しない.これにより、各数の最終結果を求める場合、スタックでこの数に影響する操作を格納することができ、1操作に遭遇するまで、または1操作に遭遇しなかった場合、初期値は入力されたa_である.i.それから問題の意味に従ってシミュレーションして解くといいです.
#include<cstdio>
#include<stack>
using namespace std;
typedef __int64 LL;
const int N = 1e5 + 10;
struct opertion
{
int t, l, r;
LL x;
}o[N];
LL a[N];
LL gcd(LL x, LL y)
{
while(y) {
LL r = x % y;
x = y;
y = r;
}
return x;
}
int main()
{
int T, n, i, j, Q;
scanf("%d",&T);
while(T--) {
scanf("%d",&n);
for(i = 1; i <= n; i++)
scanf("%I64d",&a[i]);
scanf("%d",&Q);
for(i = 0; i < Q; i++)
scanf("%d%d%d%I64d",&o[i].t, &o[i].l, &o[i].r, &o[i].x);
for(i = 1; i <= n; i++) {
stack<LL> s;
int flag = 0;
for(j = Q - 1; j >= 0; j--) {
if(i >= o[j].l && i <= o[j].r) {
s.push(o[j].x);
if(o[j].t == 1) {
flag = 1;
break;
}
}
}
if(!flag) // 1
s.push(a[i]);
while(s.size() > 1) {
LL ans = s.top(); s.pop();
LL tmp = s.top(); s.pop();
if(ans > tmp)
ans = gcd(ans, tmp);
s.push(ans);
}
printf("%I64d ", s.top());
}
printf("
");
}
return 0;
}