UVa 1588

//UVa 1588 - Kickdown
//一次循环，分类讨论
//有BUG，，，不要看了，，仅供收藏。。。欢迎大神提出意见。。。
#include<stdio.h>
int main(){
	freopen("x3.11.in","r",stdin);   
	//freopen("x3.11.out","w",stdout);
//DATE IN
	char ch, must = 1;			//ch为临时输入,must处理掩码得到0101
	bool a[2][100];				//用于存放每次的2组数据
	int count = 0, is = 0, lo[2],ca = 0; 	//count记录字符串长度并用lo[]存储，ca为答案个数
	while((ch = getchar()) != EOF){
		if(ch == '
'){
			lo[is++] = count;
			putchar('
');
			//printf("size:%d
",lo[is-1]);
			count = 0;
		}else {
			if(ch == ' ')continue;
			else{
				a[is][count] =  !(ch & must);
				printf("%d ",a[is][count]);
				count++;
			}
		}
		
	//DATE DO(此处计算并输出答案，并在最后重置is)
		if(is == 2){
			int re , m = 0, iss = 0,i = lo[0];
			//re 为重复，m向右移，iss 为开关,第i次
			while(1){
				if(i > 0)i--;else m++;
				for(int j = 0; ; j++){
					if(!m && (a[0][i+j]+a[1][j+m])==2)break;
					if(!m && (i+j)==(lo[0]-1)){re = j+1;break;}
					if(!i && j==(lo[0]-1)){re = lo[0];break;}
					if(!i && (j+m+1)==lo[1]){re >= lo[1]?iss=1:re = lo[1]-m;break;}
					if(!i && m==lo[1]){re = 0;break;}
				}
				if(re==lo[0] || re==lo[1] || iss==1 || re == 0){
					printf("case%d:%d
",++ca,lo[0]+lo[1]-re);
					break;
				}
			}
		//continue
			is = 0;
		}
	}
	return 0;
}

/*
UVa1588测试数据
case1:10
case2:4
case3:7
case4:8
case5:10
case6:8
case7:15
case8:
*/

以下为测试数据：

3.11.in

2 2 1 2 1 1 2
2 1 1 2 1 1 2 1 1 2
2 2
2 1 1 2
2 2 2 1
2 2 2 1
2 2 2 2
2 1 1 1 1 2
2 1 1 2 1 1 2 1 1 2
2212112
1 2 1 2 1 2 1 2
2 1 2 1 2 1 2 1 
2 2 1 1 2 2 1 1 2 2 
2 1 2 1 2

//AC1
#include<stdio.h>
#include<string.h>
const int maxn=100;
char a[maxn+1],b[maxn+1];
int n1,n2;
int min(const int &i,const int &j){
    return i<j?i:j;
}
int minLen(char *s1,char *s2,int &n){
	int sumLen=n1+n2, minn=min(n1,n2), len=sumLen;
	for(int i = 0; i < n; i++){
		int ok = 1, fix = min(n-i,minn);
		for(int j=0; j < fix; j++) 
			if(s1[i+j]=='2'&&s2[j]=='2'){
				ok=0; break;
			}
        if(ok&&len>sumLen-fix)len=sumLen-fix;
    }
    return len;
}
int main(){
    while(scanf("%s%s",&a,&b)==2){
        n1=strlen(a),n2=strlen(b);
        printf("%d
",min(minLen(a,b,n1),minLen(b,a,n2)));
    }
}

//AC2
#include<stdio.h>
#include<string.h>
char a[110],b[110];
int main(){
    while(scanf("%s%s",a,b) != EOF){
        int i,j,t,n1,n2;
        int temp,len1,len2;
        i=j=t=0;
        n1=strlen(a);
        n2=strlen(b);
		//
        while(j<n1 && i<n2){
            if(a[j]+b[i]-96<=3)i++,j++;
            else t++,i=0,j=t;
        }
        len1=n1+n2-i;
        i=j=t=0;
		//
        while(j<n2&&i<n1){
            if(a[j]+b[i]-96<=3)i++,j++;
            else t++,i=0,j=t;
        }
        len2=n1+n2-i;
        printf("%d
",len1 < len2 ? len1 : len2 );
    }
    return 0;
}

以下为测试数据2：