UVa 11340 Newspaper

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

Input

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

Output

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

Sample Input

1
7
a3
W 10
A 100
, 10
k7
.3
I 13
7
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Sample Output

3.74$ 

题目描述：字符有数值，给一串文本，求字符数值之和。因文件较大，使用getchar()读入。注意C++标准中char可能是unsigned也可能是signed。这题的字符需要使用unsigned char，不然会WA。运行时间30ms。

/*PROG: UVa11340*/#include <cstdio>#include <cstring>#include <cctype>using namespace std;int val[300];typedef long long LL;#define DEBUG 0#define LOG(...) do { if (DEBUG) fprintf(stderr, __VA_ARGS__); } while(0)int main(void) {int Z; scanf("%d", &Z);for (int z = 0; z < Z; ++z) {int n; scanf("%d", &n);getchar();memset(val, 0, sizeof(val));for (int i = 0; i < n; ++i) {unsigned char ch = getchar();int v; scanf("%d", &v);val[ch] = v;getchar();}LOG("val done\n");int m; scanf("%d", &m);getchar();LL ans = 0;while (m) {unsigned char ch = getchar();LOG("%c", ch);ans += val[ch];if (ch == '\n') --m;}printf("%lld.%02lld$\n", ans/100, ans%100);}return 0;}