UVA - 10361 Automatic Poetry

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UVA - 10361 Automatic Poetry

题目大意：题目大意：给出两个序列，第一个 s1s3s5，第二个 S…。输出两个序列，第一个删除括号，即 s1s2s3s4s5，第二个将 s2 与 s4 交换位置后替换…，即 Ss4s3s2s5。

解题思路：根据 < 和 > 将字符串分成 5个部分交换 s2 s4 即可

#include<iostream>#include<cstdio>#include<string.h>using namespace std;char a[1005], b[1005];char s1[1005], s2[1005], s3[1005], s4[1005], s5[1005];int main() {    int T;    scanf("%d", &T);    getchar();    while (T--) {        gets(a);        gets(b);        int len1 = strlen(a);        int len2 = strlen(b);        for (int i = 0,j = 0; i < len1; i++){            for (j = 0; a[i] != '<'; i++, j++)                s1[j] = a[i]; s1[j] = '\0'; i++;            for (j = 0; a[i] != '>'; i++, j++)                s2[j] = a[i]; s2[j] = '\0'; i++;            for (j = 0; a[i] != '<'; i++, j++)                s3[j] = a[i]; s3[j] = '\0'; i++;            for (j = 0; a[i] != '>'; i++, j++)                s4[j] = a[i]; s4[j] = '\0'; i++;            for (j = 0; a[i] != '\0'; i++, j++)                s5[j] = a[i]; s5[j] = '\0'; i++;        }    printf("%s%s%s%s%s\n", s1, s2, s3, s4, s5);    int i = 0;    while (b[i] != '.')        printf("%c", b[i++]);    printf("%s%s%s%s\n", s4, s3, s2, s5);    }    return 0;}

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