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POJ 2260 Error Correction (模拟)

发布时间:2016-12-3 6:13:53 编辑:www.fx114.net 分享查询网我要评论
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Error Correction http://poj.org/problem?id=2260 Time Limit: 1000MS Memory Limit: 65536K Description A boolean matrix has the parity property when each row and each column has an even sum, i.e. contains an even number of bits which are set. Here's a 4 x 4 matrix which has the parity property:  1 0 1 0 0 0 0 0 1 1 1 1 0 1 0 1 The sums of the rows are 2, 0, 4 and 2. The sums of the columns are 2, 2, 2 and 2.  Your job is to write a program that reads in a matrix and checks if it has the parity property. If not, your program should check if the parity property can be established by changing only one bit. If this is not possible either, the matrix should be classified as corrupt.  Input The input will contain one or more test cases. The first line of each test case contains one integer n (n<100), representing the size of the matrix. On the next n lines, there will be n integers per line. No other integers than 0 and 1 will occur in the matrix. Input will be terminated by a value of 0 for n. Output For each matrix in the input file, print one line. If the matrix already has the parity property, print "OK". If the parity property can be established by changing one bit, print "Change bit (i,j)" where i is the row and j the column of the bit to be changed. Otherwise, print "Corrupt". Sample Input 4 1 0 1 0 0 0 0 0 1 1 1 1 0 1 0 1 4 1 0 1 0 0 0 1 0 1 1 1 1 0 1 0 1 4 1 0 1 0 0 1 1 0 1 1 1 1 0 1 0 1 0 Sample Output OK Change bit (2,3) Corrupt water.只需证明:更改一个位的充要条件是行列恰各有一行是奇数。 完整代码: /*0ms,160KB*/ #include <cstdio> #include <cstring> int a[103][103], row[103], column[103]; int main(void) { int n; int i, j, k, fr, fc; while (scanf("%d", &n), n) { ///初始化 memset(row, 0, sizeof(row)); memset(column, 0, sizeof(column)); for (i = 1; i <= n; i++) { for (j = 1; j <= n; j++) { scanf("%d", &a[i][j]); row[i] += a[i][j]; column[j] += a[i][j]; } } j = 0;///检查是否为偶数 for (i = 1; i <= n; i++) { if (row[i] & 1) { ++j; fr = i; } } k = 0;///检查是否为偶数 for (i = 1; i <= n; i++) { if (column[i] & 1) { ++k; fc = i; } } if (j == 0 && k == 0) puts("OK"); else if (j == 1 && k == 1) printf("Change bit (%d,%d)\n", fr, fc); else puts("Corrupt"); } return 0; }

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