大数模余
题目描述：
A lot of people have played fortune Test(or called RP Test) before. Let ’ s create a new Test below.
Suppose the worth of a=1, b=4,c=9 … and z=26^2. Then abc can describe as 149,and abd describe as
1416. As it is too large ,we take it mod 101 as ours fortune worth. S o abd has only 2 RP. Now I give you
a name, please tell me the worth of it.
输入
The first line of the input contains the number of test cases in the file. Each test case that follows
consists of one lines. each case contains only one string s specifying a person ’ s name, which only
contains lower-case .
输出
For each test case, print a line contains the answer.
样例输入
1
a
样例输出
1
#include <iostream>#include <cstdio>#include <cstring>#include <stack>using namespace std;const int maxn = 10000;char str[maxn];int a[maxn];int T;void translate(){ stack<int> s; int point = 0; int len = strlen(str); int x; for(int i = 0; i<len; i++){ int mid = str[i]-'a'+1; int tmp = mid * mid; while(tmp) { ///把数值X分解并存入数组a中。 x = tmp%10; s.push(x); tmp = tmp/10; } int counter = (int)s.size(); for(int j=1; j<=counter; j++) { a[point++] = s.top(); s.pop(); } }}int work() { int ans = 0; for(int i = 0; a[i]>0; i++) { ans = (ans*10+a[i])%101; } return ans;}int main(){ scanf("%d", &T); while(T--) { scanf("%s", str); getchar(); translate(); int res = work(); printf("%d\n", res); memset(a, 0, sizeof(a)); memset(str, 0, sizeof(str)); } return 0;}