学步园

一题模拟题，主要要注意细节的处理。

yb牛写一遍就过了，，，，ym.....

自己wa了一个晚上。。。。。

具体方法看注释就懂了。。。

#include <stdlib.h><br /> #include <stdio.h><br /> #include <string.h></p> <p>typedef __int64 TYPE;<br /> TYPE xx[15];<br /> char str[10005], s[10005];<br /> void get(int x)<br /> {<br /> __int64 i, t = x;<br /> xx[0] = 1;<br /> for(i = 1; i<=10; i++)<br /> {<br /> xx[i] = t;<br /> t *= x;<br /> }<br /> }</p> <p>TYPE solve()<br /> {<br /> TYPE ans, c1, c2, n;<br /> int i, len = strlen(s);<br /> int op = 0;//+ 1 | - 2<br /> bool flag, flag2;<br /> i = 0, c1 = 0, c2 = 0, n = 0, ans = 0, flag = false;<br /> while(i<=len)<br /> {<br /> if( s[i] == '+' || s[i] == '-' || s[i] == 0x00)<br /> {<br /> c2 = xx[n];<br /> //根据操作符进行运算<br /> if(op == 1)<br /> {<br /> ans += (c1 * c2);<br /> flag = false;<br /> //printf( "ans:%I64d/n", c1*c2);<br /> }<br /> if(op == 2)<br /> {<br /> ans -= (c1 * c2);<br /> flag = false;<br /> //printf( "ans:-%I64d/n", c1*c2);<br /> }<br /> //首先读取操作符<br /> if(s[i] == '+') op = 1;<br /> if(s[i] == '-') op = 2;<br /> i++;<br /> c1 = 0, c2 = 0, n = 0;<br /> } else {<br /> //读取X前面的数字<br /> if(s[i] >= '0' && s[i] <= '9')<br /> {<br /> flag = true;<br /> c1 = c1*10 + s[i] - '0';<br /> i++;<br /> }</p> <p> if(s[i] == 'X')<br /> {<br /> //前缀为空<br /> if(c1 == 0 && flag == false)<br /> c1 = 1;<br /> i++;<br /> //X的后缀为空<br /> if (s[i] != '^')<br /> n = 1;<br /> }</p> <p> if(s[i] == '^')<br /> {<br /> i++;<br /> //读取^后面的数字<br /> while(1)<br /> {<br /> if(s[i] == '+' || s[i] == '-' || s[i] == 0x00)<br /> break;<br /> n = n*10 + s[i] - '0';<br /> i++;<br /> }<br /> }<br /> }<br /> }<br /> //printf("#%d/n", ans);<br /> return ans;<br /> }<br /> /*<br /> *<br /> */<br /> int main()<br /> {<br /> int x;<br /> while(scanf("%d%s", &x, str) != EOF)<br /> {<br /> get(x);<br /> if(str[0] != '-')<br /> {<br /> //所有字符串前缀都以符号开始，比较好处理<br /> s[0] = '+';<br /> strcpy(&s[1], str);<br /> } else {<br /> strcpy(s, str);<br /> }<br /> printf("%I64d/n", solve());<br /> }<br /> return 0;<br /> }

Polynomial Problem<br /> Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)<br /> Total Submission(s): 216 Accepted Submission(s): 101</p> <p>Problem Description<br /> We have learned how to obtain the value of a polynomial when we were a middle school student. If f(x) is a polynomial of degree n, we can let</p> <p>If we have x, we can get f(x) easily. But a computer can not understand the expression like above. So we had better make a program to obtain f(x).</p> <p>Input<br /> There are multiple cases in this problem and ended by the EOF. In each case, there are two lines. One is an integer means x (0<=x<=10000), the other is an expression means f(x). All coefficients ai(0<=i<=n,1<=n<=10,-10000<=ai<=10000) are integers. A correct expression maybe likes<br /> 1003X^5+234X^4-12X^3-2X^2+987X-1000</p> <p>Output<br /> For each test case, there is only one integer means the value of f(x).</p> <p>Sample Input<br /> 3<br /> 1003X^5+234X^4-12X^3-2X^2+987X-1000</p> <p>Sample Output<br /> 264302</p> <p>Notice that the writing habit of polynomial f(x) is usual such as<br /> X^6+2X^5+3X^4+4X^3+5X^2+6X+7<br /> -X^7-5X^6+3X^5-5X^4+20X^3+2X^2+3X+9<br /> X+1<br /> X^3+1<br /> X^3<br /> -X+1 etc. Any results of middle process are in the range from -1000000000 to 1000000000.<br />