UVa 11340 Newspaper (water ver.)

11340 - Newspaper

Time limit: 1.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2315

News agency pays money for articles according to some rules. Each character has its own value (some characters may have value equals to zero). Author gets his payment as a sum of all character's values in the article. You have to determine the amount of money that news agency must pay to an author.

INPUT:

The first line contains integer N (0 < N <= 5), it is a number of tests. Each test describes an integer K (0 < K <= 100), the number of paid characters. On next K lines there are table of paid characters and its values (character values are written in cents). If character can not be found in the table, then its value is equal to zero. Next, there is integer M (0 < M <= 150000). Next M lines contain an article itself. Each line can be up to 10000 characters length. Be aware of a large input size, the whole input file is about 7MB.

OUTPUT:

For each test print how much money publisher must pay for an article in format "x.yy$". Where "x" is a number of dollars without leading zeros, and "yy" number of cents with one leading zero if necessary. Examples: "3.32$", "13.07$", "71.30$", "0.09$".

SAMPLE INPUT:

17a 3W 10A 100, 10k 7. 3I 137ACM International Collegiate Programming Contest (abbreviatedas ACM-ICPC or just ICPC) is an annual multi-tiered competitionamong the universities of the world. The ICPC challenges studentsto set ever higher standards of excellence for themselvesthrough competition that rewards team work, problem analysis,and rapid software development.From Wikipedia.

SAMPLE OUTPUT:

3.74$

直接开个130大小的数组把char对应的值存起来~

/*0.036s*/#include<cstdio>#include<cstring>int ch[130];char str[10005];int main(){int n, k, val, m, len, i;long long sum;char c;scanf("%d", &n);while (n--){memset(ch, 0, sizeof(ch));scanf("%d\n", &k);while (k--){c = getchar();scanf("%d\n", &val);ch[c] = val;}scanf("%d\n", &m);sum = 0;while (m--){gets(str);len = strlen(str);for (i = 0; i < len; ++i)sum += ch[str[i]];}printf("%.2f$\n", (double)sum / 100);}return 0;}