@@ -33,9 +33,9 @@ public class Code02_MagicGirl1 {
3333 public static int [] v = new int [MAXN ];
3434
3535 // pos[1..cntp]是当前序列块的下标
36- // que [1..cntq]是整包结算的查询编号
36+ // qid [1..cntq]是整包结算的查询编号
3737 public static int [] pos = new int [MAXN ];
38- public static int [] que = new int [MAXN ];
38+ public static int [] qid = new int [MAXN ];
3939 public static int cntp ;
4040 public static int cntq ;
4141
@@ -81,7 +81,7 @@ public static void merge(int i, int curPre, int curSuf, int curLen, int curAns)
8181 }
8282
8383 // 整包结算
84- // que [1..cntq]是查询编号,每条查询整包[l..r]
84+ // qid [1..cntq]是查询编号,每条查询整包[l..r]
8585 // 根据序列块的数字状况,更新每个查询的答案信息
8686 public static void calc (int l , int r ) {
8787 for (int i = l ; i <= r ; i ++) {
@@ -90,10 +90,10 @@ public static void calc(int l, int r) {
9090 next [i ] = i + 1 ;
9191 }
9292 radix (pos , arr , cntp );
93- radix (que , v , cntq );
93+ radix (qid , v , cntq );
9494 int curPre = 0 , curSuf = 0 , curLen = r - l + 1 , curAns = 0 ;
9595 for (int i = 1 , j = 1 , idx ; i <= cntq ; i ++) {
96- while (j <= cntp && arr [pos [j ]] <= v [que [i ]]) {
96+ while (j <= cntp && arr [pos [j ]] <= v [qid [i ]]) {
9797 idx = pos [j ];
9898 if (last [idx ] == l - 1 ) {
9999 curPre += next [idx ] - idx ;
@@ -106,7 +106,7 @@ public static void calc(int l, int r) {
106106 next [last [idx ]] = next [idx ];
107107 j ++;
108108 }
109- merge (que [i ], curPre , curSuf , curLen , curAns );
109+ merge (qid [i ], curPre , curSuf , curLen , curAns );
110110 }
111111 cntp = cntq = 0 ;
112112 }
@@ -120,7 +120,7 @@ public static void compute(int l, int r) {
120120 }
121121 } else {
122122 if (x [qi ] <= l && r <= y [qi ]) {
123- que [++cntq ] = qi ;
123+ qid [++cntq ] = qi ;
124124 } else {
125125 for (int i = Math .max (x [qi ], l ); i <= Math .min (y [qi ], r ); i ++) {
126126 if (arr [i ] <= v [qi ]) {
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