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1 /* 2 题意：表达式转换 3 模拟：题目不难，也好理解题意，就是有坑！具体的看测试样例。。。 4 */ 5 #include 6 #include 7 #include 8 #include 9 #include 10 #include 11 #include 12 #include 13 #include 14 #include 15 #include 16 #include 17 using namespace std; 18 19 const int MAXN = 1e4 + 10; 20 const int INF = 0x3f3f3f3f; 21 22 char s[] = { 'p', 'q', 'r', 'u', 'v', 'w', 'x', 'y', 'z'}; 23 24 int main(void) //HDOJ 5095 Linearization of the kernel functions in SVM 25 { 26 //freopen ("F.in", "r", stdin); 27 28 int t; scanf ("%d", &t); 29 while (t--) 30 { 31 int a[11]; bool first = false; int j = -1; 32 for (int i=1; i<=10; ++i) scanf ("%d", &a[i]); 33 for (int i=1; i<=10; ++i) 34 { 35 j++; 36 if (!a[i]) continue; 37 if (a[i] > 0 && first) putchar ('+'); 38 39 if (a[i] == -1) 40 { 41 if (i < 10) putchar ('-'); 42 else printf ("-1"); 43 } 44 else 45 { 46 if (a[i] == 1) 47 { 48 if (i == 10) putchar ('1'); 49 } 50 else printf ("%d", a[i]); 51 } 52 53 first = true; 54 if (j <= 8) printf ("%c", s[j]); 55 } 56 if (!first) printf ("0"); 57 puts (""); 58 } 59 60 return 0; 61 } 62 63 /* 64 21 65 0 46 3 4 -5 -22 -8 -32 24 27 66 2 31 -5 0 0 12 0 0 -49 12 67 0 1 0 0 0 0 0 0 0 -1 68 0 0 0 0 0 0 0 0 0 0 69 0 0 0 0 0 0 0 0 0 1 70 1 -1 1 -1 1 1 1 1 1 0 71 0 46 3 4 -5 -22 -8 -32 24 27 72 2 31 -5 0 0 12 0 0 -49 12 73 0 0 0 0 0 0 0 0 0 0 74 1 2 3 4 5 6 7 8 9 10 75 -1 2 3 -5 8 3 0 6 9 20 76 1 -1 1 1 1 -1 -1 -1 0 0 77 10000 123 123 123 456 12354 123 12345 45 10110 78 0 0 0 1 2 3 -1 -2 -3 56 79 0 0 0 -1 -2 0 0 1 23 45 80 */ 81 82 /* 83 46q+3r+4u-5v-22w-8x-32y+24z+27 84 2p+31q-5r+12w-49z+12 85 q-1 86 0 87 1 88 p-q+r-u+v+w+x+y+z 89 46q+3r+4u-5v-22w-8x-32y+24z+27 90 2p+31q-5r+12w-49z+12 91 0 92 p+2q+3r+4u+5v+6w+7x+8y+9z+10 93 -p+2q+3r-5u+8v+3w+6y+9z+20 94 p-q+r+u+v-w-x-y 95 10000p+123q+123r+123u+456v+12354w+123x+12345y+45z+10110 96 u+2v+3w-x-2y-3z+56 97 -u-2v+y+23z+45 98 -u-2v+y+23z+45 99 -u-2v+y+23z+45100 -u-2v+y+23z+45101 -u-2v+y+23z+45102 -u-2v+y+23z+45103 -u-2v+y+23z+45104 */
转载于:https://www.cnblogs.com/Running-Time/p/4514897.html