P1320压缩技术（续集版

感觉这题还是蛮难的对我来说，通过这题我才知道原来字符串输入不碰到空格就会一起输进来
我参考了一写题解自己又写了自己的解法，vs中的scanf_s和scanf()用法不太一样，之前按scanf写法写一直在报错，心态有点被搞崩了
在这里插入图片描述

下面是洛谷的解法

#include <stdio.h>
#include <math.h>
#include <string.h>
int n;//存储点阵是几行几列
int a[200][2];//用来存储每一位
int now = 0;//用来储存现在是第几位数
char x[200];
char y[200];
int m;
int k;//储存上一位的y值
int many = 0;//存储一共有几个数
int f;

int main() {
	scanf("%s", &x);
	if (x[0] == '1') m = 1;
	//int p = 1;
	while (x[many] == '0'|| x[many]=='1') {
		k = m;
		m = x[many] - '0';
		many++;
		if (k == 1 && m == 0) now++;
		a[now][m]++;
	}
	
	/*while (scanf_s("%s", y, 1)) {
		m = y - "0";
		a[now][m]++;
		k = m;
		//m = 3;
		//scanf_s("%d", &y);
		many++;
		if (k == 1 && m == 0) now++;
	}*/
	for (int l = 1; l < many; l++) {
		
			scanf("%s", &y);
			for (int o = 0; o < many; o++) {
				k = m;
				m = y[o] - '0';
				if (k == 1 && m == 0) now++;
				a[now][m]++;
			}
			//m = 3;
			//scanf_s("%d", &y);
			//many++;
	}
	//f = many;
	printf("%d ", many);
	for (int i=0; i <= now; i++) {
		for (int j=0; j < 2; j++) {
			if(i==now&&a[i][j]==0&&a[i][j+1]==0)break;
             printf("%d ", a[i][j]);
		}

	}

}

下面是vs中的写法

#include <stdio.h>
#include <math.h>
#include <string.h>
int n;//存储点阵是几行几列
int a[200][2];//用来存储每一位
int now = 0;//用来储存现在是第几位数
char x[200];
char y[200];
int m;
int k;//储存上一位的y值
int many = 0;//存储一共有几个数
int f;

int main() {
	scanf_s("%s", x,200);
	if (x[0] == '1') m = 1;
	//int p = 1;
	while (x[many] == '0'|| x[many]=='1') {
		k = m;
		m = x[many] - '0';
		many++;
		if (k == 1 && m == 0) now++;
		a[now][m]++;
	}
	
	/*while (scanf_s("%s", y, 1)) {
		m = y - "0";
		a[now][m]++;
		k = m;
		//m = 3;
		//scanf_s("%d", &y);
		many++;
		if (k == 1 && m == 0) now++;
	}*/
	for (int l = 1; l < many; l++) {
		
			scanf_s("%s", y, 200);
			for (int o = 0; o < many; o++) {
				k = m;
				m = y[o] - '0';
				if (k == 1 && m == 0) now++;
				a[now][m]++;
			}
			//m = 3;
			//scanf_s("%d", &y);
			//many++;
	}
	//f = many;
	printf("%d ", many);
	for (int i=0; i <= now; i++) {
		for (int j=0; j < 2; j++) {
			if (i == now && a[i][2]==0)break;
			printf("%d ", a[i][j]);
		}

	}

}