codeforces 754A(思维题)
2017-05-14 11:44
363 查看
One spring day on his way to university Lesha found an array
A. Lesha likes to split arrays into several parts. This time Lesha decided to split the array
A into several, possibly one, new arrays so that the sum of elements in each of the new arrays is not zero. One more condition is that if we place the new arrays one after another they will form the old array
A.
Lesha is tired now so he asked you to split the array. Help Lesha!
Input
The first line contains single integer n (1 ≤ n ≤ 100) — the number of elements in the array
A.
The next line contains n integers
a1, a2, ..., an ( - 103 ≤ ai ≤ 103) —
the elements of the array A.
Output
If it is not possible to split the array A and satisfy all the constraints, print single line containing "NO" (without quotes).
Otherwise in the first line print "YES" (without quotes). In the next line print single integer
k — the number of new arrays. In each of the next
k lines print two integers
li and
ri which denote the subarray
A[li...
ri] of the initial array
A being the i-th new array. Integers
li,
ri should satisfy the following conditions:
l1 = 1
rk = n
ri + 1 = li + 1 for each
1 ≤ i < k.
If there are multiple answers, print any of them.
Example
Input
Output
Input
Output
Input
Output
Input
Output
/*题意让你将一个数组任意拆分为多个数组,按照原来的元素的顺序,满足的条件是每组数组的和不能为0,这些数组元素必须是原来的顺序一个接一个,可能会有多种解人以输出一个即可*/
/*思路:最简单的方法就是将原来的数组拆分为的数组只含有一个元素,保证每个元素不为零,需要特判,可能原来数组有多个0,都为0,前半部分为0,后半部分为0,中间部分为0的情况。详情见代码*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int num[200];
int main()
{
int n;
while(scanf("%d", &n) != EOF){
memset(num, 0, sizeof(num));
int cnt = 0;
for(int i = 1; i <= n; i++){
scanf("%d", &num[i]);
if(num[i] == 0) ++cnt;
}
int flag = 0;
int f0 = 0;
if(cnt == n){ //按照这种思路输出为no的情况只有所有元素为0(cnt统计的是0的个数)
printf("NO\n");
continue;
}
else{
printf("YES\n");
int t = n - cnt;
printf("%d\n", t);
for(int i = 1; i <= n; i++){
if(t == 1) { //这里表示当前只有一个数字不为零其他都为0那么只用输出一组即可
printf("%d %d\n", i, n);
break;
}
else {
if(num[i] == 0){
if(!f0){ //若当前为0,直到下一个非0时才能停止那么需要标记
printf("%d ", i);
f0 = 1;
}
flag = 1;
continue;
}
if(flag){
printf("%d\n", i);
flag = 0;
f0 = 0;
t--; //这里的原因是为了解决若有多个0
continue;
}
else{
printf("%d %d\n", i, i);
t--;
}
}
}
}
}
return 0;
}
A. Lesha likes to split arrays into several parts. This time Lesha decided to split the array
A into several, possibly one, new arrays so that the sum of elements in each of the new arrays is not zero. One more condition is that if we place the new arrays one after another they will form the old array
A.
Lesha is tired now so he asked you to split the array. Help Lesha!
Input
The first line contains single integer n (1 ≤ n ≤ 100) — the number of elements in the array
A.
The next line contains n integers
a1, a2, ..., an ( - 103 ≤ ai ≤ 103) —
the elements of the array A.
Output
If it is not possible to split the array A and satisfy all the constraints, print single line containing "NO" (without quotes).
Otherwise in the first line print "YES" (without quotes). In the next line print single integer
k — the number of new arrays. In each of the next
k lines print two integers
li and
ri which denote the subarray
A[li...
ri] of the initial array
A being the i-th new array. Integers
li,
ri should satisfy the following conditions:
l1 = 1
rk = n
ri + 1 = li + 1 for each
1 ≤ i < k.
If there are multiple answers, print any of them.
Example
Input
3 1 2 -3
Output
YES 2 1 2 3 3
Input
8 9 -12 3 4 -4 -10 7 3
Output
YES 2 1 2 3 8
Input
1 0
Output
NO
Input
4 1 2 3 -5
Output
YES 4 1 1 2 2 3 3 4 4
/*题意让你将一个数组任意拆分为多个数组,按照原来的元素的顺序,满足的条件是每组数组的和不能为0,这些数组元素必须是原来的顺序一个接一个,可能会有多种解人以输出一个即可*/
/*思路:最简单的方法就是将原来的数组拆分为的数组只含有一个元素,保证每个元素不为零,需要特判,可能原来数组有多个0,都为0,前半部分为0,后半部分为0,中间部分为0的情况。详情见代码*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int num[200];
int main()
{
int n;
while(scanf("%d", &n) != EOF){
memset(num, 0, sizeof(num));
int cnt = 0;
for(int i = 1; i <= n; i++){
scanf("%d", &num[i]);
if(num[i] == 0) ++cnt;
}
int flag = 0;
int f0 = 0;
if(cnt == n){ //按照这种思路输出为no的情况只有所有元素为0(cnt统计的是0的个数)
printf("NO\n");
continue;
}
else{
printf("YES\n");
int t = n - cnt;
printf("%d\n", t);
for(int i = 1; i <= n; i++){
if(t == 1) { //这里表示当前只有一个数字不为零其他都为0那么只用输出一组即可
printf("%d %d\n", i, n);
break;
}
else {
if(num[i] == 0){
if(!f0){ //若当前为0,直到下一个非0时才能停止那么需要标记
printf("%d ", i);
f0 = 1;
}
flag = 1;
continue;
}
if(flag){
printf("%d\n", i);
flag = 0;
f0 = 0;
t--; //这里的原因是为了解决若有多个0
continue;
}
else{
printf("%d %d\n", i, i);
t--;
}
}
}
}
}
return 0;
}
相关文章推荐
- Codeforces 554B:Ohana Cleans Up(思维+水题)
- CodeForces 11B Jumping Jack(思维)
- codeforces-597【思维】
- Codeforces - 612D. The Union of k-Segments 排序+思维
- Codeforces 660D Number of Parallelograms 【思维】
- CodeForces - 786B Legacy(线段树 +最短路+思维好题)
- CodeForces - 821C Okabe and Boxes(思维)
- Codeforces 557C Arthur and Table【思维】
- Codeforces 296C Greg and Array【思维】
- Codeforces 371B - Filya and Homework(思维)
- Codeforces 747D Persistent Bookcase【Dfs+思维】好题!好题!
- Codeforces 591E Three States【优先队列Bfs+思维】
- codeforces 466B Wonder Room(思维,暴力)
- Codeforces 320A - Raising Bacteria(思维)
- CodeForces 633B A Trivial Problem(思维,阶乘 0 的个数)
- CodeForces 628B New Skateboard 思维
- codeforces 501 C. Misha and Forest (思维)
- CodeForces - 757C - Felicity is Coming!(思维)
- CodeForces 672D (思维+二分)
- CodeForces - 729D Sea Battle(思维题)