zoj3829 Known Notation --- 2014 ACM-ICPC Asia Mudanjiang Regional Contest
2014-10-17 17:24
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根据规则可以发现,一个*前面必须至少要有2个数字,一个(11*)这样的又可以当成一个数字,
所以总的数字的个数一定大于*数,不然就要添加数字。
添加数字后,可以保证一定可以通过交换得到解,那么肯定是*往后放,数字往前放,交换到符合条件为止。
所以总的数字的个数一定大于*数,不然就要添加数字。
添加数字后,可以保证一定可以通过交换得到解,那么肯定是*往后放,数字往前放,交换到符合条件为止。
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; char s[1010]; int len,ans,opp[1010],num[1010],op,cnt; bool check(int b) { int a=0; for(int i=0;i<len;i++) { if(s[i]=='*') a++; else b++; if(a>=b) return 0; } return 1; } int main() { int T,i,j,b; scanf("%d",&T); while(T--) { scanf("%s",s); len=strlen(s); cnt=op=ans=0; for(i=0;i<len;i++) { if(s[i]=='*') opp[op++]=i; else num[cnt++]=i; } if(op==0) { puts("0"); continue; } for(i=0,j=cnt-1;i<j;i++,j--) swap(num[i],num[j]); if(op>=cnt) ans+=(op-cnt+1); int fir=0,last=0; b=ans; while(!check(b)) { ans++; swap(s[opp[fir++]],s[num[last++]]); } printf("%d\n",ans); } return 0; }
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