HDU-6140 Killer Names - 2017 Multi-University Training Contest - Team 8(思维)
2017-08-17 20:14
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题意:
给定一个序列,序列的每个值都有属性,N代表可加可减,L代表只能加,D代表只能减,当然这些都可以用或者不用,并且这个序列的每个值还有一个奇怪的式子进行限制。给定一个k,问k是否可能用这个序列构成。
思路:
没做出来,想明白就很简单。。。
首先a1 = 1, b1 = N 构成[-1, 1]的所有值,然后根据i >= 2时ai <= 奇怪的式子,bi为'L'时其实就是通过区间左值来限制ai不能大于它,bi为'D'时其实就是通过区间右值来限制ai不能大于它,从而根据题解的保证(一组只能构成区间[L, R]内的数再组合上一个绝对值小于L绝对值的负值,则必能构成区间[L-该负值, R]所有的数。同理,再组合上一个小于R的正值,则必能构成区间[L,
R+该正值]所有的数),i从小到大去扩展一个区间即可,最后再判断一次k是否属于该区间。
代码:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3+5;
int a[maxn];
char b[maxn];
int main()
{
int t, n, k, L, R;
scanf("%d", &t);
while(t--)
{
scanf("%d %d", &n, &k);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
for(int i = 1; i <= n; ++i) scanf(" %c", &b[i]);
L = R = 0;
for(int i = 1; i <= n; ++i)
{
if(b[i] == 'N') L -= a[i], R += a[i];
if(b[i] == 'L') L -= a[i];
if(b[i] == 'D') R += a[i];
}
if(k >= L && k <= R) puts("yes");
else puts("no");
}
return 0;
}
继续加油~
给定一个序列,序列的每个值都有属性,N代表可加可减,L代表只能加,D代表只能减,当然这些都可以用或者不用,并且这个序列的每个值还有一个奇怪的式子进行限制。给定一个k,问k是否可能用这个序列构成。
思路:
没做出来,想明白就很简单。。。
首先a1 = 1, b1 = N 构成[-1, 1]的所有值,然后根据i >= 2时ai <= 奇怪的式子,bi为'L'时其实就是通过区间左值来限制ai不能大于它,bi为'D'时其实就是通过区间右值来限制ai不能大于它,从而根据题解的保证(一组只能构成区间[L, R]内的数再组合上一个绝对值小于L绝对值的负值,则必能构成区间[L-该负值, R]所有的数。同理,再组合上一个小于R的正值,则必能构成区间[L,
R+该正值]所有的数),i从小到大去扩展一个区间即可,最后再判断一次k是否属于该区间。
代码:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3+5;
int a[maxn];
char b[maxn];
int main()
{
int t, n, k, L, R;
scanf("%d", &t);
while(t--)
{
scanf("%d %d", &n, &k);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
for(int i = 1; i <= n; ++i) scanf(" %c", &b[i]);
L = R = 0;
for(int i = 1; i <= n; ++i)
{
if(b[i] == 'N') L -= a[i], R += a[i];
if(b[i] == 'L') L -= a[i];
if(b[i] == 'D') R += a[i];
}
if(k >= L && k <= R) puts("yes");
else puts("no");
}
return 0;
}
继续加油~
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